Math 380 Elementary Algebra Textbook Spring 2012 Edition Department of Mathematics College of the Redwoods January 13, 2012 Copyright All parts of this prealgebra textbook are copyrighted © 2011 in the name of the Department of Mathematics, College of the Redwoods. They are not in the public domain. However, they are being made available free for use in educational institutions. This offer does not extend to any application that is made for profit. Users who have such applications in mind should contact David Arnold at david-arnold@redwoods.edu or Bruce Wagner at bruce-wagner@redwoods.edu. This work is licensed under the Creative Commons Attribution-Non- Commercial-NoDerivs 3.0 Unported License. To view a copy of this li- cense, visit http://creativecommons.Org/licenses/by-nc-nd/3.0/ or send a letter to Creative Commons, 171 Second Street, Suite 300, San Fran- cisco, California, 94105, USA. Contents The Arithmetic of Numbers 1 1.1 An Introduction to the Integers 2 The Integers 3 Absolute Value 4 Integer Addition 5 Mathematical Properties of Addition 6 Integer Subtraction 7 Integer Multiplication 8 Mathematical Properties of Multiplication 9 Exponents 10 Graphing Calculator: Negating versus Subtracting 11 Exercises 14 Answers 15 1.2 Order of Operations 16 Grouping Symbols 19 Absolute Value Bars as Grouping Symbols 20 Nested Grouping Symbols 20 Evaluating Algebraic Expressions 21 Evaluating Fractions 22 Using the Graphing Calculator 23 Exercises 27 Answers 30 1.3 The Rational Numbers 31 Reducing Fractions to Lowest Terms 31 Multiplying Fractions 33 Dividing Fractions 35 Adding Fractions 36 Order of Operations 37 Fractions on the Graphing Calculator 39 Exercises 43 Answers 46 CONTENTS 1.4 Decimal Notation 48 Adding and Subtracting Decimals 49 Multiplication and Division of Decimals 51 Order of Operations 52 Rounding Using the Graphing Calculator 55 Exercises 57 Answers 59 1.5 Algebraic Expressions 60 The Distributive Property 61 Speeding Things Up a Bit 62 Distributing a Negative Sign 63 Combining Like Terms 64 Order of Operations 66 Exercises 69 Answers 70 Solving Linear Equations and Inequalities 73 2.1 Solving Equations: One Step 74 Equivalent Equations 74 Wrap and Unwrap, Do and Undo 75 Operations that Produce Equivalent Equations 76 More Operations That Produce Equivalent Equations .... 79 Writing Mathematics 82 Exercises 84 Answers 85 2.2 Solving Equations: Multiple Steps 87 Variables on Both Sides of the Equation 90 Simplifying Expressions When Solving Equations 92 Exercises 96 Answers 97 2.3 Clearing Fractions and Decimals 99 Canceling is More Efficient 100 Clearing Fractions from an Equation 101 Clearing Decimals from an Equation 106 Exercises 109 Answers 110 2.4 Formulae Ill Clearing Fractions 113 Geometric Formulae 115 Exercises 118 Answers 120 2.5 Applications 121 Exercises 132 Answers 134 2.6 Inequalities 135 Ordering the Real Numbers 136 CONTENTS Set-Builder Notation 136 Interval Notation 138 Equivalent Inequalities 139 Reversing the Inequality Sign 141 Multiple Steps 143 Summary Table of Set-Builder and Interval Notation 146 Exercises 147 Answers 149 Introduction to Graphing 151 3.1 Graphing Equations by Hand 152 The Cartesian Coordinate System 152 Plotting Ordered Pairs 154 Equations in Two Variables 156 Graphing Equations in Two Variables 157 Guidelines and Requirements 159 Using the TABLE Feature of the Graphing Calculator .... 160 Exercises 165 Answers 168 3.2 The Graphing Calculator 171 Reproducing Calculator Results on Homework Paper .... 174 Adjusting the Viewing Window 176 Exercises 179 Answers 180 3.3 Rates and Slope 183 Measuring the Change in a Variable 184 Slope as Rate 185 The Steepness of a Line 189 The Geometry of the Slope of a Line 192 Exercises 196 Answers 199 3.4 Slope-Intercept Form of a Line 201 Applications 206 Exercises 209 Answers 212 3.5 Point-Slope Form of a Line 213 Parallel Lines 217 Perpendicular Lines 219 Applications 222 Exercises 224 Answers 225 3.6 Standard Form of a Line 228 Slope- Intercept to Standard Form 230 Point-Slope to Standard Form 232 Intercepts 234 Horizontal and Vertical Lines 237 CONTENTS Exercises 240 Answers 242 Systems of Linear Equations 245 4.1 Solving Systems by Graphing 246 Exceptional Cases 250 Solving Systems with the Graphing Calculator 253 Exercises 259 Answers 260 4.2 Solving Systems by Substitution 262 Exceptional Cases Revisited 268 Exercises 270 Answers 272 4.3 Solving Systems by Elimination 273 Exceptional Cases 276 Exercises 279 Answers 280 4.4 Applications of Linear Systems 282 Exercises 290 Answers 291 Polynomial Functions 293 5.1 Functions 294 Mapping Diagrams 296 Function Definition 297 Mapping Diagram Notation 299 Function Notation 301 Interchanging y and f{x) 303 Exercises 305 Answers 307 5.2 Polynomials 308 Ascending and Descending Powers 309 The Degree of a Polynomial 311 Polynomial Functions 312 The Graph of a Polynomial Function 313 Exercises 317 Answers 319 5.3 Applications of Polynomials 321 Zeros and ^-intercepts of a Function 325 Exercises 330 Answers 333 5.4 Adding and Subtracting Polynomials 334 Negating a Polynomial 335 Subtracting Polynomials 336 Some Applications 338 Exercises 340 CONTENTS vii Answers 342 5.5 Laws of Exponents 343 Multiplying With Like Bases 344 Dividing With Like Bases 345 Raising a Power to a Power 347 Raising a Product to a Power 349 Raising a Quotient to a Power 350 Exercises 353 Answers 355 5.6 Multiplying Polynomials 356 The Product of Monomials 356 Multiplying a Monomial and a Polynomial 357 Multiplying Polynomials 359 Speeding Things Up a Bit 360 Some Applications 362 Exercises 366 Answers 368 5.7 Special Products 370 The FOIL Method 370 The Difference of Squares 373 Squaring a Binomial 375 An Application 378 Exercises 380 Answers 382 6 Factoring 385 6.1 The Greatest Common Factor 386 Finding the Greatest Common Factor of Monomials 388 Factor Out the GCF 390 Speeding Things Up a Bit 392 Factoring by Grouping 394 Exercises 396 Answers 398 6.2 Solving Nonlinear Equations 399 Linear versus Nonlinear 400 Using the Graphing Calculator 404 Exercises 410 Answers 411 6.3 Factoring ax 2 + bx + c when a = 1 413 The ac-Method 413 Speeding Things Up a Bit 416 Nonlinear Equations Revisited 417 Exercises 424 Answers 425 6.4 Factoring ax 2 + bx + c when a^ 1 427 Speeding Things Up a Bit 428 i CONTENTS Nonlinear Equations Revisited 431 Exercises 437 Answers 438 6.5 Factoring Special Forms 440 Perfect Square Trinomials 440 The Difference of Squares 444 Factoring Completely 446 Nonlinear Equations Revisited 447 Exercises 452 Answers 454 6.6 Factoring Strategy 456 Using the Calculator to Assist the ac-Method 462 Exercises 464 Answers 465 6.7 Applications of Factoring 467 Exercises 474 Answers 475 Rational Expressions 477 7.1 Negative Exponents 478 Laws of Exponents 480 Raising to a Negative Integer 482 Applying the Laws of Exponents 484 Clearing Negative Exponents 485 Exercises 488 Answers 490 7.2 Scientific Notation 492 Multiplying Decimal Numbers by Powers of Ten 493 Scientific Notation Form 495 Placing a Number in Scientific Notation 495 Scientific Notation and the Graphing Calculator 497 Exercises 502 Answers 504 7.3 Simplifying Rational Expressions 505 Multiplying and Dividing Rational Expressions 505 Adding and Subtracting Rational Expressions 507 The Least Common Denominator 508 Dividing a Polynomial by a Monomial 511 Exercises 513 Answers 514 7.4 Solving Rational Equations 516 Solving Rational Equations with the Graphing Calculator . . 518 Numerical Applications 521 Exercises 523 Answers 524 7.5 Direct and Inverse Variation 525 CONTENTS Inversely Proportional 527 Exercises 531 Answers 533 8 Quadratic Functions 535 8.1 Introduction to Radical Notation 536 Using the Graphing Calculator 539 Approximating Square Roots 542 Exercises 545 Answers 546 8.2 Simplifying Radical Expressions 547 Simple Radical Form 548 The Pythagorean Theorem 549 Proof of the Pythagorean Theorem 550 Applications 553 Exercises 555 Answers 557 8.3 Completing the Square 558 Perfect Square Trinomials Revisited 561 Completing the Square 562 Solving Equations by Completing the Square 563 Exercises 568 Answers 569 8.4 The Quadratic Formula 571 Exercises 580 Answers 582 Index 585 Chapter 1 The Arithmetic of Numbers In 1960, Belgian Geologist Jean de Heinzelein de Braucourt discovered the Is- liango bone in central Africa. This bone, dated to be more than 20,000 years old, is believed to be the oldest known artifact indicating the use of arith- metic. The first written records indicate the Egyptians and Babylonians used arithmetic as early as 2000 BC. The Mayans used arithmetic to make astro- nomical computations, and developed the concept of zero over 2,000 years ago. The word "arithmetic" is derived from the Greek word arithmos (translated as "number" ) . It is the oldest and most elementary branch of mathematics and is used for a variety of tasks ranging from simple counting to advanced science and business calculations. CHAPTER 1. THE ARITHMETIC OF NUMBERS 1.1 An Introduction to the Integers We begin with the set of counting numbers, formally called the set of natural numbers. If we add the number zero to the set of natural numbers, then we have a set of numbers that are called the whole numbers. The Whole Numbers. The set W={0,1,2,3,4,5,. ••} is called the set of whole numbers. The number is special, in that whenever you add it to another whole number, you get the identical number as an answer. Additive Identity Property. If a is any whole number, then a + = a. For this reason, the whole number is called the additive identity. Thus, for example, 3 + = 3, 15 + = 15, and 123 + = 123. These are all examples of the additive identity property. Every natural number has an opposite, so that when you add them together, their sum is zero. Additive Inverse Property. If a is any natural number, then define the opposite of a, symbolized by —a, so that a + (-a) = 0. The number —a is called the "opposite of a," or more formally, the additive inverse of a. For example, the opposite (additive inverse) of 3 is —3, and 3 + (—3) = 0. The opposite (additive inverse) of 12 is —12, and 12 + (—12) = 0. The opposite of 1.1. AN INTROD UCTION TO THE INTEGERS 254 is —254, and 254+ (—254) = 0. These are all examples of additive inverses and the additive inverse property. Because 7 + (—7) = 0, we've said that —7 is the opposite (additive inverse) of 7. However, we can also turn that around and say that 7 is the opposite of —7. If we translate the phrase "the opposite of —7 is 7" into mathematical symbols, we get —(—7) = 7. Thus, for example, -(-11) = 11, -(-103) = 103, and -(-1255) = 1255. The Integers If we collect all the natural numbers and their additive inverses, then include the number zero, we have a collection of numbers called the integers. The Integers. The set Z = {...,-5, -4, -3, -2, -1,0,1,2,3,4,5,. ■■} is called the set of integers. The integers can be made to correspond to points on a line in a very natural manner. First, draw a line, then locate the number zero anywhere you wish. Secondly, place the number one to the right of zero. This determines the length of one unit. Finally, locate the numbers 1, 2, 3, 4, 5, . . . to the right of zero, then their opposites (additive inverses) —1, —2, —3, —4, —5, . . . to the left of zero (see Figure 1.1). <H 1 1 1 1 1 1 1 1 1 h- ► -5-4-3-2-10 1 2 3 4 5 Figure 1.1: Each integer corresponds to a unique position on the number line. Note that as we move to the right on the number line, the integers get larger. On the other hand, as we move to the left on the number line, the integers get smaller. Positive and negative integers. On the number line, some integers lie to the right of zero and some lie to the left of zero. CHAPTER 1. THE ARITHMETIC OF NUMBERS • If a is an integer. integer that lies to the right of zero , then a is called a positive • If a is an integer. integer that lies to the left of zero, then a is called s negative Thus, 4, 25, and 142 are positive integers, while —7, —53, and —435 are negative integers. Absolute Value The absolute value (or magnitude) of an integer is defined as follows. The Absolute Value of an Integer. If a is an integer, then the absolute value of a, written \a\, is defined as the distance between the integer and zero on the number line. You Try It! Simplify: | - 23 1 EXAMPLE 1. Simplify | - 4|. Solution: Consider the position of —4 on the number line. Note that —4 lies four units away from zero. 4 units -«-H H h H 1 1 h H h +->■ -1 1 Answer: I - 23| = 23 Because the absolute value (magnitude) of an integer equals its distance from zero, | - 4| = 4. □ In similar fashion: • The integer 5 lies five units away from zero. Hence, |5| = 5. • The integer lies zero units away from zero, Hence, |0| = 0. Note that the absolute value of any number is either positive or zero. That is, the absolute value of a number is nonnegative (not negative). 1.1. AN INTROD UCTION TO THE INTEGERS Integer Addition This section is designed to provide a quick review of integer addition. For a more thorough introduction to integer addition, read section two of chapter two of our prealgebra textbook, provided online at the following URL: http : //msenux . redwoods . edu/PreAlgText/contents/chapter2/ chapter2 .pdf We consider the first of two cases. Adding Integers with Like Signs. To add two integers with like signs (both positive or both negative), add their magnitudes (absolute values), then prefix their common sign. You Try It! EXAMPLE 2. Simplify 7+12. Solution: We have like signs. The magnitudes (absolute values) of 7 and 12 are 7 and 12, respectively. If we add the magnitudes, we get 19. If we prefix the common sign, we get 19. That is: 7 + 12= 19 Simplify: 13 + 28 Answer: 41 □ You Try It! EXAMPLE 3. Simplify -8+ (-9). Solution: We have like signs. The magnitudes (absolute values) of —8 and —9 are 8 and 9, respectively. If we add the magnitudes, we get 17. If we prefix the common sign, we get —17. That is: Simplify: —12 -21) + (-9) 17 Answer: -33 □ Next, we consider the case where we have unlike signs. Adding Integers with Unlike Signs. To add two integers with unlike signs (one positive and one negative), subtract the integer with the smaller magnitude (absolute value) from the number with the larger magnitude, then prefix the sign of the integer with the larger magnitude. CHAPTER 1. THE ARITHMETIC OF NUMBERS You Try It! Simplify: 12 -29) EXAMPLE 4. Simplify -14+11. Solution: We have unlike signs. The magnitudes (absolute values) of —14 and 11 are 14 and 11, respectively. If we subtract the smaller magnitude from the larger, we get 3. The number —14 has the larger magnitude, so we prefix our answer with its negative sign. That is: Answer: —17 -14+11 □ You Try It! Simplify: 32 + (-90) EXAMPLE 5. Simplify 40 + (-25). Solution: We have unlike signs. The magnitudes (absolute values) of 40 and —25 are 40 and 25, respectively. If we subtract the smaller magnitude from the larger, we get 15. The number 40 has the larger magnitude, so we prefix our answer with its positive sign. That is: Answer: -58 40 +(-25) = 15 □ Mathematical Properties of Addition The order in which we add integers does not matter. That is, —20 + 34 gives an answer identical to the sum 34 + (—20). In both cases, the answer is 14. This fact is called the commutative property of addition. Next, when we add three integers, it does not matter which two we add first. For example, if we add the second and third of three numbers first, we get: -11 + (-2 + 5) = -11 + 3 Parentheses first: Add: -11 + 3 = 1.1. AN INTRODUCTION TO THE INTEGERS 7 On the other hand, if we add the first and second of three numbers first, we get: (-11 + (-2)) + 5 = -13 + 5 Parentheses first: -11 + (-2) = -13 = -8 Add: -13 + 5 = -8 Thus, —11 + (—2 + 5) = (—11 + (—2)) + 5. This fact is called the associative property of addition. The Associative Property of Addition. If a, 6, and c are any three integers, then: a + (b + c) = {a + b) + c Integer Subtraction Subtraction is the inverse, or the opposite, of addition. You Try It! EXAMPLE 6. Simplify: -13-27 Simplify: -11-15 Solution: The "opposite" (additive inverse) of 27 is —27. So, subtracting 27 is the same as adding —27. -13 - 27 = -13 + (-27) Subtracting 27 is the same as adding —27. = —50 Add the magnitudes, then prefix the common negative sign. Answer: —26 You Try It! □ EXAMPLE 7. Simplify: -27- (-50) Simplify: -18 - (-54) Answer: 36 You Try It! Simplify: (-18)(-5) Answer: 90 8 CHAPTER 1. THE ARITHMETIC OF NUMBERS Solution: The "opposite" (additive inverse) of —50 is —(—50), or 50. So, subtracting —50 is the same as adding 50. -27- (-50) = -27 + 50 23 Subtracting —50 is the same as adding 50. Subtract the smaller magnitude from the larger magnitude, then prefix the sign of the larger magnitude. □ Integer Multiplication This section is designed to provide a quick review of multiplication and division of integers. For a more thorough introduction to integer multiplication and division, read section four of chapter two of our prealgebra textbook, provided online at the following URL: http : //msenux . redwoods . edu/PreAlgText/contents/chapter2/ chapter2 .pdf Like Signs. If a and 6 are integers with like signs (both positive or both negative), then the product ab and the quotient a/b are positive. (+)(+) = = + or (+)/(+) = + (-)(-) = = + or (")/(-) = + EXAMPLE 8. Simplify each of the following expressions: (a) (2)(3) (b) (-12)(-8) (c) -14/(-2) Solution: When multiplying or dividing, like signs yield a positive result. (a)(2)(3)=6 (b) (-12)(-8) = 96 (c) -14/(-2) = 7 □ Unlike Signs. If a and b are integers with unlike signs (one positive and one negative), then the product ab and the quotient a/b are negative. (+)(") = - or (+)/(") = - (")(+) = - or (")/(+) = - 1.1. AN INTROD UCTION TO THE INTEGERS You Try It! EXAMPLE 9. Simplify each of the following expressions: (a) (2)(-12) (b) (-9)(12) (c) 24/(-8) Solution: When multiplying or dividing, unlike signs yield a negative result, (a) (2) (-12) = -24 (b) (-9) (12) = -108 (c) 24/(-8) = -3 Simplify: (-19)(3) Answer: —57 Mathematical Properties of Multiplication The order in which we multiply integers does not matter. That is, (— 8)(5) gives an answer identical to (5) (—8). In both cases, the answer is —40. This fact is called the commutative property of multiplication. The Commutative Property of Multiplication. If a and b are any two integers, then: a ■ b = b ■ a Next, when we multiply three integers, it does not matter which two we multiply first. If we multiply the second and third of three numbers first, we get: (-3) [ (-4)(-5) ] = (-3) (20) Brackets first: (-4)(-5) = 20 = -60 Multiply: (-3)(20) = -60 On the other hand, if we multiply the first and second of three numbers first, we get: [ (-3)(-4) ] (-5) = (12) (-5) Brackets first: (-3)(-4) = 12 = -60 Multiply: (12)(-5) = -60 Thus, (— 3) [(— 4)(— 5)] = [(— 3)(— 4)] (— 5). This fact is called the associative property of multiplication. The Associative Property of Multiplication. If o, 6, and c are any three integers, then: a ■ (b ■ c) = (a ■ b) ■ c □ When you multiply an integer by 1, you get the identical number back as the product. For example, (1)(5) = 5 and ( — 11) (1) = —11. This fact is known as the multiplicative identity property. 10 CHAPTER 1. THE ARITHMETIC OF NUMBERS The Multiplicative Identity Property. If a is any integer, then: 1 ■ a = a and a ■ 1 = a For this reason, the integer 1 is called the "multiplicative identity." Finally, note that (— 1)(5) = —5. Thus, multiplying 5 by —1 is identical to taking the "opposite" of 5 or negating 5. The Multiplicative Property of — 1. Multiplying by minus one is identical to negating. That is: (— l)a = —a Exponents In the exponential expression a™, the number a is called the base, while the number n is called the exponent. We now define what is meant by an exponent. Exponents. Let a be an integer and let n be any whole number. If n ^ 0, then: a = a ■ a ■ a ■ a n times That is, to calculate a™, write a as a factor n times. You Try It! Simplify: (-2) EXAMPLE 10. Simplify (-2) 3 . Solution: In the exponential expression (— 2) 3 , note that —2 is the base, while 3 is the exponent. The exponent tells us to write the base as a factor three times. Simplify the result by performing the multiplications in order, moving from left to right. Answer: 4 (-2) 3 = (-2)(-2)(-2) (4)(-2) —2 as a factor, three times. Multiply: (-2) (-2) = 4. Multiply: (4) (-2) = -8. Thus, (-2) 2 □ In Example 10, note that the product of three negative factors is negative. Let's try another example. 1.1. AN INTROD UCTION TO THE INTEGERS 11 EXAMPLE 11. Simplify (-2) 4 . Solution: In the exponential expression (— 2) 4 , note that — 2 is the base, while 4 is the exponent. The exponent tells us to write the base as a factor four times. Simplify the result by performing the multiplications in order, moving from left to right. (-2) 4 = (-2)(-2)(-2)(-2) = (4)(-2)(-2) = (-8X-2) = 16 -2 as a factor, four times. Multiply Multiply Multiply (-2)(-2)=4. (4)(-2) = -8. (-8)(-2) = 16. You Try It! Simplify: (-2) Thus, (-2) 4 = 16. Answer: -32 In Example 11, note that the product of four negative factors is positive. Examples 10 and 11 reveal the following pattern. □ Odd or Even Exponents. 1. When a negative integer is raised to an even exponent, the result is positive. 2. When a negative integer is raised to an odd exponent, the result is neg- ative. Graphing Calculator: Negating versus Subtracting Consider the view of the lower half of the TI84 graphing calculator in Figure 1.2. Note that there are two keys that contain some sort of negative sign, one on the bottom row of keys, and another in the last column of keys on the right, positioned just above the plus symbol. (-) and Figure 1.2: Lower half of the TI-84. The first of these buttons is the unary "negation" operator. If you want to negate a single (thus the word "unary") number, then this is the key to use. For example, enter -3 by pressing the following button sequence. The result is shown in Figure 1.3. EDH ENTER 12 CHAPTER 1. THE ARITHMETIC OF NUMBERS The second button is the binary "subtraction" operator. If you want to subtract one number from another number (thus the word "binary" ) , then this is the key to use. For example, enter 7-15 by pressing the following button sequence. The result is shown in Figure 1.4. QH00 ENTER Figure 1.3: Negating a number. Figure 1.4: Subtract two numbers. Important Point. Do not interchange the roles of the unary negation oper- ator and the binary subtraction operator. 1. To negate a number, use: (-) 2. To subtract one number from another, use: If you interchange the roles of these operators, the calculator will respond that you've made a "syntax error" (see Figures 1.5 and 1.6). ERR: SYNTAX IBQuit ZSGota Figure 1.5: Using the wrong symbol for subtraction. Figure 1.6: The resulting syntax er- ror. You Try It! Use the graphing calculator to evaluate (-225) 3 . EXAMPLE 12. Use the TI84 graphing calculator to simplify each of the following expressons: (a) -717-432 (b) (232)(-313) (c) (-17) 3 1.1. AN INTROD UCTION TO THE INTEGERS 13 Solution: The minus sign in each of these examples looks exactly the same, but sometimes it is used as a "negative" sign and sometimes it is used as a "subtraction" sign. a) The expression —717 — 432 asks us to subtract 432 from "negative" 717. Enter the following sequence of keystrokes to produce the result shown in the first image in Figure 1.7. EDmmmBsss ENTER Hence, -717-432 -1149. b) The expression (232) (—313) asks us to find the product of 232 and "nega- tive" 313. Enter the following sequence of keystrokes to produce the result shown in the second image in Figure 1.7. HmSHEDHmS ENTER Hence, (232)(-313) = -72616. c) The expression (— 17) 3 asks us to raise "negative" to the third power. Enter the following sequence of keystrokes to produce the result shown in the third image in Figure 1.7. The "caret" symbol BUB is located just above the division key in the rightmost column of the TI84 graphing calculator. Eomm s ENTER Hence, (-17) s -4913. -717-432 -1149 232* -313 -72616 (-17> A 3 -4913 Figure 1.7: Calculations made on the graphing calculator. Answer: -11390625 □ 14 CHAPTER 1. THE ARITHMETIC OF NUMBERS f> t* t* Exercises ■** •** ■** In Exercises 1-8, simplify each of the following expressions. 1. |5|. 5. |2|. 2. |1|. 6. |8|. 3. |-2|. 7. |-4|. 4. |-1|. 8. |-6|. In Exercises 9-24, simplify each of the following expressions as much as possible. 9. -91 + (-147) 17. 37 +(-86) 10. -23 + (-13) 18. 143 + (-88) 11.96 + 145 19. 66 +(-85) 12. 16 + 127 20. 33 + (-41) 13. -76 + 46 21. 57 + 20 14.-11 + 21 22.66 + 110 15. -59 + (-12) 23.-48+127 16. -40 +(-58) 24.-48 + 92 In Exercises 25-32, find the difference. 25. -20- (-10) 29.-77-26 26. -20 -(-20) 30.-96-92 27. -62-7 31. -7- (-16) 28.-82-62 32. -20 -(-5) In Exercises 33-40, compute the exact value. 33. (-8) 6 36. (-4) 6 34. (-3) 5 37. (-9) 2 35. (-7) 5 38. (-4) 2 1.1. AN INTROD UCTION TO THE INTEGERS 15 39. (-A) 4 40. (-5) 4 In Exercises 41-52, use your graphing calculator to compute the given expression. 41. -562- 1728 42. -3125- (-576) 43. -400- (-8225) 44. -8176 + 578 45. (-856)(232) 46. (-335)(-87) 47. (-815)(-357 48. (753)(-9753) 49. (-18) 3 50. (-16) 4 51. (-13) 5 52. (-15) 6 £*• £*< £*■ Answers -m •&§ >*& 1. 5 3. 2 5. 2 7. 4 9. -238 11. 241 13. -30 15. -71 17. -49 19. -19 21. 77 23. 79 25. -10 27. -69 29. -103 31. 9 33. 262144 35. -16807 37. 81 39. 256 41. -2290 43. 7825 45. -198592 47. 2916885 49. -5832 51. -371293 16 CHAPTER 1. THE ARITHMETIC OF NUMBERS 1.2 Order of Operations The order in which we evaluate expressions can be ambiguous. Take, for ex- ample, the expression —4 + 2-8. If we perform the addition first, then we get — 16 as a result (the question mark over the equal sign indicates that the result is questionable). -4 + 2-8 = -2-8 ? = -16. On the other hand, if we perform the multiplication first, then we get 12 as a result. -4 + 2-8 = -4 + 16 = 12. So, what are we to do? Of course, grouping symbols would remove the ambiguity. Grouping Symbols. Parentheses, brackets, and absolute value bars can be used to group parts of an expression. For example: 3 + 5(9-11) or -2- [-2-5(1-3)] or 6 - 3| - 3 - 4| In each case, the rule is "evaluate the expression inside the grouping sym- bols first." If the grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols first. Thus, if the example above is grouped as follows, we are forced to evaluate the expression inside the parentheses first. (-4 + 2) • 8 = -2 ■ 8 Parentheses first: -4 + 2 = -2 = -16 Multiply: -2 ■ 8 = -16 Another way to avoid ambiguities in evaluating expressions is to establish an order in which operations should be performed. The following guidelines should always be strictly enforced when evaluating expressions. Rules Guiding Order of Operations. When evaluatinj 5 expressions, pro- ceed in the following order. 1. Evaluate expressions contained in grouping symbols first. If grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols first. 1.2. ORDER OF OPERATIONS 17 2. Evaluate all exponents that appear in the expression. 3. Perform all multiplications and divisions in the order that they appear in the expression, moving left to right. 4. Perform all additions and subtractions in the order that they appear in the expression, moving left to right. EXAMPLE 1. Simplify: -3-4-8 Solution: Because of the established Rules Guiding Order of Operations, this expression is no longer ambiguous. There are no grouping symbols or exponents present, so we immediately go to rule three, evaluate all multiplications and divisions in the order that they appear, moving left to right. After that we invoke rule four, performing all additions and subtractions in the order that they appear, moving left to right. You Try It! Simplify: -4 + 2 -3-4- -3-32 -3 + (-32) -35 Multiply first: 4 ■ 8 Add the opposite. Add: -3 +(-32) = 32 -35 Thus, -3-4 -35. Answer: 12 □ Writing Mathematics. When simplifying expressions, observe the following rule to neatly arrange your work One equal sign per line. This means that you should not arrange your work horizontally. -2-4- (- -8) = -2-(- -32) = - -2 + 32 = 30 That's three equal signs on a sinj de line. Rather, arrange your work vertically, keeping equal signs aligned in a i :olumn. - 2-4 (-8) = -2-(- -32) -2 + 32 30 18 CHAPTER 1. THE ARITHMETIC OF NUMBERS You Try It! Simplify: -24/(-3)(2) Answer: 16 EXAMPLE 2. Simplify: 54/(-9)(2) Solution: There are no grouping symbols or exponents present, so we imme- diately go to rule three, evaluate all multiplications and divisions in the order that they appear, moving left to right. 54/(-9)(2) -6(2) -12 Divide first: 54/ (-9) = -6 Multiply: -6(2) = -12 Thus, 54/(-9)(2) = -12. □ Example 2 can be a source of confusion for many readers. Note that multipli- cation takes no preference over division, nor does division take preference over multiplication. Multiplications and divisions have the same level of preference and must be performed in the order that they occur, moving from left to right. If not, the wrong answer will be obtained. Warning! Here is what happens if you perform the multiplication in Exam- ple 2 before the division. 54/(-9)(2) = 54/(-18) = -3 Multiply: (-9) (2) Divide: 54/ (-18) = -18 This is incorrect! Multiplications and divisions must be performed in the order that they occur, moving from left to right. You Try It! Simplify: — 15^ EXAMPLE 3. Simplify: (a) (-7) 2 and (b) -7 2 Solution. Recall that for any integer a, we have (— l)a = —a. Because negat- ing is equivalent to multiplying by —1, the Rules Guiding Order of Operations require that we address grouping symbols and exponents before negation. a) Because of the grouping symbols, we negate first, then square. That is, (-7) 2 = (-7)(-7) = 49. b) There are no grouping symbols in this example. Thus, we must square first, then negate. That is, -72 Answer: -225 Thus, (-7) 2 = 49, but -7 2 (— 7) 2 is different from — 7 2 . -(7-7) = -49. -49. Note: This example demonstrates that D 1.2. ORDER OF OPERATIONS 19 Let's try an example that has a mixture of exponents, multiplication, and subtraction. EXAMPLE 4. Simplify: -3 - 2(-4) 2 Solution. The Rules Guiding Order of Operations require that we address exponents first, then multiplications, then subtractions. -4) 2 -3-2(16) -3-32 -3 +(-32) -35 Exponent first: (—4) = Multiply: 2(16) = 32 Add the opposite. Add: -3 +(-32) = -35 16 Thus, -3-2(-4) 2 -35. You Try It! Simplify: 4(-2) 2 Answer: 27 □ Grouping Symbols The Rules Guiding Order of Operations require that expressions inside group- ing symbols (parentheses, brackets, or curly braces) be evaluated first. EXAMPLE 5. Simplify: -2(3 - 4) 2 + 5(1 - 2) 3 Solution. The Rules Guiding Order of Operations require that we first eval- uate the expressions contained inside the grouping symbols. -2(3 -4) 2 + 5(1 -2) 3 = -2(3 + (-4)) 2 + 5(1 + (-2)) 3 Add the opposites. = -2(-l) 2 + 5(-l) 3 Parentheses first: 3 + (-4) = -1 and 1 + (-2) = -1. You Try It! Simplify: -2-3(-2-3) s Evaluate the exponents next, perform the multiplications, then add = -2(l) + 5(-l) = -2 + (-5) Exponents: (—1) =1 and (-1) 3 = -1. Multiply: -2(1) = -2 and 5(-l) = -5. Add: -2 +(-5) = -7 Thus, -2(3-4)2 + 5(1-2)= Answer: 373 □ 20 CHAPTER 1. THE ARITHMETIC OF NUMBERS You Try It! Simplify: 4-61 Absolute Value Bars as Grouping Symbols Like parentheses and brackets, you must evaluate what is inside them first, then take the absolute value of the result. EXAMPLE 6. Simplify: -8 - |5 - 11| Solution. We must first evaluate what is inside the absolute value bars. — 8 — |5 — 11| = -8 - |5 + (-11)| Add the opposite. = -8 - | - 6| Add: 5 + (-11) = -6. The number —6 is 6 units from zero on the number line. Hence, I — 6| = 6. Answer: —10 Thus, = -8-6 Add: | - 6| = 6. = -8 + (-6) Add the opposite = -14 Add. 5- - Ill = -14. □ You Try It! Simplify: -2-2[-2-2(-2-2)] Nested Grouping Symbols When grouping symbols are nested, the Rules Guiding Order of Operations tell us to evaluate the innermost expressions first. EXAMPLE 7. Simplify: -3 - 4[-3 - 4(-3 - 4)] Solution. The Rules Guiding Order of Operations require that we first ad- dress the expression contained in the innermost grouping symbols. That is, we evaluate the expression contained inside the brackets first. -3 - 4[-3 - 4(-3 - 4)] = -3 - 4[-3 - 4(-3 + (-4))] Add the opposite. = -3 - 4[-3 - 4(-7)] Add: -3 + (-4) = -7 Next, we evaluate the expression contained inside the brackets. -3-4[-3- (-28)] -3 - 4[-3 + 28] -3 - 4[25] Multiply: 4(-7) = -28 Add the opposite. Add: -3 + 28 = 25 Now we multiply, then subtract. 1.2. ORDER OF OPERATIONS 21 = -3- 100 = -3 +(-100) = -103 Thus, -3 - 4[-3 - 4(-3 - 4)] Multiply: 4{25} = 100 Add the opposite. Add: -3 +(-100) = -103 -103. Answer: -14 □ Evaluating Algebraic Expressions Variable. A variable is a symbol (usually a letter) that stands for an unknown value that may vary. Let's add the definition of an algebraic expression. Algebraic Expression. When we combine numbers and variables in a valid way, using operations such as addition, subtraction, multiplication, division, exponentiation, the resulting combination of mathematical symbols is called an algebraic expression. Thus, 2a, and y being formed by a combination of numbers, variables, and mathematical oper- ators, are valid algebraic expressions. An algebraic expression must be well-formed. For example, 2 + -5a; is not a valid expression because there is no term following the plus sign (it is not valid to write H — with nothing between these operators). Similarly, 2 + 3(2 is not well-formed because parentheses are not balanced. In this section we will evaluate algebraic expressions for given values of the variables contained in the expressions. Here are some simple tips to help you be successful. Tips for Evaluating Algebraic Expressions. 22 CHAPTER 1. THE ARITHMETIC OF NUMBERS 1. Replace all occurrences of variables in the expression with open paren- theses. Leave room between the parentheses to substitute the given value of the variable. 2. Substitute the given values of variables in the open parentheses prepared in the first step. 3. Evaluate the resulting expression according to the Rules Guiding Order of Operations. You Try It! If X: evaluate x 3 - 2 and y = — 1 , y 3 EXAMPLE 8. Evaluate the expression x 2 — 2xy + y 2 at x = —3 and y = 2. Solution. Following Tips for Evaluating Algebraic Expressions, first replace all occurrences of variables in the expression x 2 — Ixy + y 2 with open parentheses. Next, substitute the given values of variables (—3 for x and 2 for y) in the open parentheses. x 2 -2xy + y 2 = ( ) 2 -2( )( ) + ( ) 2 = (-3) 2 -2(-3)(2) + (2) 2 Finally, follow the Rules Guiding Order of Operations to evaluate the resulting expression. 2xy + y -( ) 2 "2( )() + () 2 : (-3) 2 -2(-3)(2) + (2) 2 = 9-2(-3)(2)+4 = 9-(-6)(2) + 4 9- (-12) + 4 : 9+ 12 + 4 :25 Original expression. Replace variables with parentheses. Substitute —3 for x and 2 for y. Evaluate exponents first. Left to right, multiply: 2(— 3) = —6 Left to right, multiply: (-6) (2) = -12 Add the opposite. Add. Answer: Thus, if x = —3 and y = 2, then x 2 — 2xy + y 2 25. □ Evaluating Fractions If a fraction bar is present, evaluate the numerator and denominator separately according to the Rules Guiding Order of Operations, then perform the division in the final step. 1.2. ORDER OF OPERATIONS 23 You Try It! EXAMPLE 9. Evaluate the expression ad — be a + b at a = 5, b = —3, c = 2, and d = —4. Solution. Following Tips for Evaluating Algebraic Expressions, first replace all occurrences of variables in the expression (ad — be)/ (a + b) with open paren- theses. Next, substitute the given values of variables (5 for a, —3 for b, 2 for c, and —4 for d) in the open parentheses. ad-be _ {){ )-( )( ) a+b ()+( ) (5)(-4) - (-3)(2) (5) + (-3) Finally, follow the Rules Guiding Order of Operations to evaluate the result- ing expression. Note that we evaluate the expressions in the numerator and denominator separately, then divide. ad-bc ( )( )-( )( ) a + b ( ) + ( ) = (5)(-4)- -(" -3)(2) (5) + (" 3) -20- (- -6) 2 -20 + 6 2 -14 2 = -7 Thus, if « = 5, b = - -3, c = 2, Replace variables with parentheses. Substitute: 5 for a, —3 for 6, 2 for c, —4 for d Numerator: (5)(-4) = -20, (-3)(2) = -6 Denominator: 5 + (— 3) = 2 Numerator: Add the opposite Numerator: —20 + 6 = —14 Divide. 2, and d = -4, then (ad - be)/ (a + b) = -7. If a = and d -7, b = —3, c = —15, - — 14, evaluate: Answer: □ Using the Graphing Calculator The graphing calculator is a splendid tool for evaluating algebraic expressions, particularly when the numbers involved are large. 24 CHAPTER 1. THE ARITHMETIC OF NUMBERS You Try It! Use the graphing calculator to evaluate -2-2[-2-2(-2-2)]. EXAMPLE 10. Use the graphing calculator to simplify the following expres- sion. -213 - 35[-18 - 211(15 - 223)] Solution. The first difficulty with this expression is the fact that the graphing calculator does not have a bracket symbol for the purposes of grouping. The calculator has only parentheses for grouping. So we first convert our expression to the following: -213 - 35(-18 - 211(15 - 223)) Note that brackets and parentheses are completely interchangeable. The next difficulty is determining which of the minus signs are negation symbols and which are subtraction symbols. If the minus sign does not appear between two numbers, it is a negation symbol. If the minus sign does appear between two numbers, it is a subtraction symbol. Hence, we enter the following keystrokes on our calculator. The result is shown in Figure 1.8. ED0QSB00DBEDS0BH QQBiQEI000 ENTER Answer: —14 -2 13-35* (-18-211 *U5-223>> -1535663 Figure 1.8: Calculating -213 - 35[-18 - 211(15 - 223)]. Thus, -213 -35[-18- 211(15 -223)] = -1,535,663. □ You Try It! Use the graphing calculator to evaluate 10 + 10 10 + 10 EXAMPLE 11. Use the graphing calculator to evaluate 5 + 5 5 + 5 1.2. ORDER OF OPERATIONS 2o 5 + 5 10 5 + 5 " 10 = 1 Solution. You might ask "Why do we need a calculator to evaluate this exceedingly simple expression?" After all, it's very easy to compute. Simplify numerator and denominator. Divide: 10/10 = 1. Well, let's enter the expression 5+5/5+5 in the calculator and see how well we understand the Rules Guiding Order of Operations (see first image in Figure 1.9). Whoa! How did the calculator get 11? The answer is supposed to be 1! Let's slow down and apply the Rules Guiding Order of Operations to the expression 5+5/5+5. 5 + 5/5 + 5 = 5+ - + 5 5 =5+1+5 o = 11 Add: 5 + 1 + 5 = Aha! That's how the calculator got 11. 5 + 5/5 + 5 is equivalent to Let's change the order of evaluation by using grouping symbols. Note that: ivide first. ivide: — = 5 1. dd: 5 + 1 + 5 -1 + 5 (5 + 5)/(5 + 5) = 10/10 = 1 Parentheses first. Divide: 10/10= 1. That is: (5 + 5)/(5 + 5) is equivalent to 5 + 5 Enter (5+5) (5+5) and press the ENTER key to produce the output shown in the second image in Figure 1.9. 5+5/5+5 11 (5+5VC5+5) Figure 1.9: Calculating 5 + 5 5 + 5' Answer: 1 □ 26 CHAPTER 1. THE ARITHMETIC OF NUMBERS You Try It! Use the graphing calculator to evaluate |o — b\ at a = -312 and b = -875. 9*a ;'0 Figure 1.10: Upper half of the TI84. The graphing calculator has memory locations available for "storing" values. They are lettered A-Z and appear on the calculator case, in alphabetic order as you move from left to right and down the keyboard. Storing values in these memory locations is an efficient way to evaluate algebraic expressions containing variables. Use the key to access these memory locations. EXAMPLE 12. Use the graphing calculator to evaluate \a\ and b = -875. at a -312 Solution. First store —312 in the variable A with the following keystrokes. To select the letter A, press the ALPHA key, then the MATH key, located in the upper left-hand corner of the calculator (see Figure 1.10). EDHQE ALPHA ENTER Next, store —875 in the variable B with the following keystrokes. To select the letter B, press the ALPHA key, then the APPS key. 1000 ALPHA The results of these keystrokes are shown in the first image in Figure 1.11. Now we need to enter the expression \a\ — \b\. The absolute value function is located in the MATH menu. When you press the MATH key, you'll notice submenus MATH, NUM, CPX, and PRB across the top row of the MATH menu. Use the right-arrow key to select the NUM submenu (see the second image in Figure 1.11). Note that abs( is the first entry on this menu. This is the absolute value function needed for this example. Enter the expression abs(A)-abs(B) as shown in the third image in Figure 1.11. Use the ALPHA key as described above to enter the variables A and B and close the parentheses using the right parentheses key from the keyboard. Press the ENTER key to evaluate your expression. -312-»FI -875-»B -312 -S75 MATH [Bill CPX PRE iHabs< 2: round ( 3: iPartC 4: f PartC 5: int( 6: min< 7-1-maxt Figure 1.11: Evaluate \a\ — \b\ at a = —312 and b -875. Answer: 563 Thus, \a\ -563. □ 10. -(-30) 11. -3 5 12. -3 2 13. 48 -r 4(6) 14. 96 -=-6(4) 15. -52-8(-8 16. -8-7(-3) 17. ("2) 4 18. ("4) 4 1.2. ORDER OF OPERATIONS 27 ».**.**• Exercises ■** * «•* In Exercises 1-18, simplify the given expression. 1. -12 + 6(-4) 2. 11 + 11(7) 3. -(-2) 5 4. -(-5) 3 5. — | — 40| 6. -|-42| 7. -24/(-6)(-l) 8. 45/(-3)(3) 9. -(-50) In Exercises 19-42, simplify the given expression. 19. 9-3(2) 2 31.-6-5(4-6) 20. -4-4(2) 2 32. -5-5(-7-7) 21. 17— 10|13 — 14| 33. 9 + (9- 6) 3 - 5 22. 18 — 3| — 20 — 5| 34. 12 + (8 -3) 3 -6 23. -4 + 5(-4) 3 35. -5 + 3(4) 2 24. 3 + 3(-4) 3 36.2 + 3(2) 2 25. 8-5(-l-6) 37. 8- (5 - 2) 3 + 6 26. 8-4(-5-5) 38. 9- (12- ll) 2 +4 27. (10 - 8) 2 - (7 - 5) 3 39. |6 - 15| - | - 17 - 11| 28. (8 - 10) 2 - (4 - 5) 3 40. | - 18 - 19| - | - 3 - 12| 29. 6 - 9(6 - 4(9 - 7)) 41. 5 - 5(5 - 6(6 - 4)) 30. 4 - 3(3 - 5(7 - 2)) 42. 4 - 6(4 - 7(8 - 5)) 28 CHAPTER 1. THE ARITHMETIC OF NUMBERS In Exercises 43-58, evaluate the expression at the given values of x and y 43. 4a; 2 + 3xy + Ay 2 at x = -3 and y = 44. 3a; 2 - 3xy + 2y 2 at x = 4 and y = -3 45. -8x + 9 at x = -9 46. -12a; + 10 at x = 2 47. — 5x 2 + 2xy — 4y 2 at a; = 5 and y = 48. 3a; 2 + 3xy - by 2 at x = and y = 3 49. 3a; 2 + 3a; - 4 at a; = 5 50. 2a; 2 + 6x - 5 at a; = 6 51. -2a; 2 + 2y 2 at a; = 1 and y = -2 52. -5a; 2 + 5y 2 at x = -4 and y = 53. -3a; 2 - 6a; + 3 at x = 2 54. -7a; 2 + 9x + 5 at x = -7 55. — 6a; — 1 at a; = 1 56. 10a; + 7 at x = 9 57. 3a; 2 - 2y 2 at a: = -3 and y = -2 58. -3a; 2 + 2y 2 at x = 2 and y = 2 59. Evaluate a 2 + 6 2 a + 6 at a = 27 and 6 = -30. 60. Evaluate a 2 + b 2 a + b at a = —63 and b = 77. 61. Evaluate a + b c— d at a = -42, b = 25, c = 26, and d = 43. 62. Evaluate a + 6 c — d at a = 38, b = 42, c = 10, and d = 50. 63. Evaluate a — b cd at a = —7, b = 48, c = 5, and d = 11. 64. Evaluate cd at a = -46, 6 = 46, c = 23, and d = 2. 65. Evaluate the expressions a 2 + b 2 and (a + 6) 2 at a = 3 and 6 = 4. Do the expressions produce the same results? 66. Evaluate the expressions a 2 b 2 and (ab) 2 at a = 3 and 6 = 4. Do the expressions produce the same results? 67. Evaluate the expressions |a||6| and \ab\ at a = —3 and 6 = 5. Do the expressions produce the same results? 68. Evaluate the expressions \a\ + \b\ and \a + b\ at a = —3 and 6 = 5. Do the expressions produce the same results? 1.2. ORDER OF OPERATIONS 2!) In Exercises 69-72, use a graphing calculator to evaluate the given expression. 69. -236-324(-576 + 57) — 270-900 70. -443 + 27(-414-22) 71. 72. 300 - 174 3000 - 952 144 - 400 73. Use a graphing calculator to evaluate the expression a a at a = —93 and b = 84 by first storing -93 in the variable A and 84 in the variable B, then entering the expression (A~2+B~2)/ (A+B) . .2 , ia 74. Use a graphing calculator to evaluate the expression a a + b at a = — 76 and b = 77 by first storing -76 in the variable A and 77 in the variable B, then entering the expression (A~2+B~2)/ (A+B) . 75. The formula 9 C + 32 will change a Celsius temperature to a Fahrenheit temperature. Given that the Celsius temperature is C = 60° C, find the equivalent Fahrenheit temperature. 76. The surface area of a cardboard box is given by the formula S = 2WH + 2LH + 2LW, where W and L are the width and length of the base of the box and H is its height. If W = 2 centimeters, L = 8 centimeters, and H = 2 centimeters, find the surface area of the box. 77. The kinetic energy (in joules) of an object having mass m (in kilograms) and veloc- ity v (in meters per second) is given by the formula K = —mv . 2 Given that the mass of the object is m = 7 kilograms and its velocity is v = 50 meters per second, calculate the kinetic energy of the object. 78. The area of a trapezoid is given by the formula A = -(b 1 + b 2 ) h, where b± and 62 are the lengths of the parallel bases and h is the height of the trapezoid. If the lengths of the bases are 21 yards and 11 yards, respectively, and if the height is 22 yards, find the area of the trapezoid. 30 CHAPTER 1. THE ARITHMETIC OF NUMBERS s*- j*- £*■ Answers <#$ >#s >*$ 1. -36 41. 40 3. 32 43. 36 5. -40 45. 81 7 - - 4 47. -125 9. 50 49 86 11. -243 13. 72 15. 12 17. 16 19. -3 21. 7 23. -324 25. -27 27. -4 29. 24 31. 4 33. 31 35. 43 37. -13 39. -19 51. 6 53. -21 55. -7 57. 19 59. -543 61. 1 63. -1 65. No. 67. Yes. 69. 167920 71. -5 73. -8 75. 140° F 77. 8750 joules 1.3. THE RATIONAL NUMBERS 31 1.3 The Rational Numbers We begin with the definition of a rational number. Rational Numbers. Any number that can be expressed in the form where p and q are integers, q 7^ 0, is called a rational number. The letter used to represent the set of rational numbers. That is: p/q, Qis Q-- = < — : p and U q are integers, q 7^ > Because —2/3, 4/5, and 123/(— 12) have the form p/q, where p and q are integers, each is an example of a rational number. If you think you hear the word "fraction" when we say "rational number," you are correct in your thinking. Any number that can be expressed as a fraction, where the numerator and denominator are integers, is a rational number. Every integer is also a rational number. Take, for example, the integer — 12. There are a number of ways we can express —12 as a fraction with integer numerator and denominator, —12/1, 24/ (—2), and —36/3 being a few. Reducing Fractions to Lowest Terms First, we define what is meant by the greatest common divisor of two integers. The Greatest Common Divisor. Given two integers a and b, the greatest common divisor of a and 6 is the largest integer that divides evenly (with no remainder) into both a and b. The notation GCD(a, b) is used to represent the greatest common divisor of a and b. For example, GCD(12, 18) = 6, GCD(32, 40) = 8, and GCD(18, 27) = 9. We can now state when a fraction is reduced to lowest terms. Lowest Terms. A fraction a/b is said to be reduced to lowest terms if and only if GCD(a,6) = 1. A common technique used to reduce a fraction to lowest terms is to divide both numerator and denominator by their greatest common denominator. EXAMPLE 1. Reduce 8/12 to lowest terms. You Try It! Reduce: -48/60 32 CHAPTER 1. THE ARITHMETIC OF NUMBERS Answer: —4/5 Solution: Note that GCD(8, 12) = 4. Divide both numerator and denomina- tor by 4. -r4 12 12 -=-4 _ 2 ~ 3 Thus, 8/12 = 2/3. Divide numerator and denominator by GCD(8,12) = 4. Simplify numerator and denominator. □ Recall the definition of a prime number. Prime Number. A natural number greater than one is prime if and only if its only divisors are one and itself. You Try It! Reduce 18/24 to lowest terms. For example, 7 is prime (its only divisors are 1 and 7), but 14 is not (its divisors are 1, 2, 7, and 14). In like fashion, 2, 3, and 5 are prime, but 6, 15, and 21 are not prime. EXAMPLE 2. Reduce 10/40 to lowest terms. Solution: Note that GCD(10,40) = 10. Divide numerator and denominator by 10. Divide numerator and denominator by GCD(10,40) = 10. Simplify numerator and denominator. 10 10- -10 40 40- 1 ~ 4 -10 Alternate solution: Use factor trees to express both numerator and denom- inator as a product of prime factors. 10 >D 40 / \ 4 10 2 2 Hence, 10 = 2 ■ 5 and 40 = 2 • 2 • 2 • 5. Now, to reduce 10/40 to lowest terms, replace the numerator and denominator with their prime factorizations, then cancel factors that are in common to both numerator and denominator. 1.3. THE RATIONAL NUMBERS :W 10 40 2-5 2-2-2-5 1 Prime factor numerator and denominator. Cancel common factors. Simplify numerator and denominator. When we cancel a 2 from both the numerator and denominator, we're actually dividing both numerator and denominator by 2. A similar statement can be made about canceling the 5. Canceling both 2 and a 5 is equivalent to dividing both numerator and denominator by 10. This explains the 1 in the numerator when all factors cancel. Answer: 3/4 Example 2 demonstrates an important point. When all factors cancel. When all of the factors cancel in either numerator or denominator, the resulting numerator or denominator is equal to one. □ Multiplying Fractions First, the definition. Multiplication of Fractions. If a/b and c/d are two fractions, then their product is defined as follows: a c ac 1"d~ bd Thus, to find the product of a/b and c/d, simply multiply numerators and multiply denominators. For example: 1 3 2 ' 4 and 2 7 5 ' 3 14 15 and 5_ 48 Like integer multiplication, like signs yield a positive answer, unlike signs yield a negative answer. Of course, when necessary, remember to reduce your answer to lowest terms. EXAMPLE 3. Simplify: 14 10 "20 ' 21 You Try It! Simplify: 27 20 34 CHAPTER 1. THE ARITHMETIC OF NUMBERS Solution: Multiply numerators and denominators, then reduce to lowest terms. Multiply numerators and denominators. Prime factor. Cancel common factors. Simplify. Note that when all the factors cancel from the numerator, you are left with a Answer: 6/5 1. Thus, (-14/20) • (10/21) = -1/3. □ 14 10 20 ' 21 ~ 140 ~420 2-2-5-7 2-2-3-5-7 = 1 ~3 Cancellation Rule. When multiplying fractions, cancel common factors ac- cording to the following rule: "Cancel a factor in a numerator for an identical factor in a denominator." You Try It! Simplify: _6_ 45 35 14 The rule is "cancel something on the top for something on the bottom." Thus, an alternate approach to multiplying fractions is to factor numerators and denominators in place, then cancel a factor in a numerator for an identical factor in a denominator. EXAMPLE 4. Simplify: 15 14 ~9~ Solution: Factor numerators and denominators in place, then cancel common factors in the numerators for common factors in the denominators. 15 t) 3-5 2-2-2 / 2-7 2-5 2-2-2 ( 2-7 V #'3 35 ~12 Factor numerators and denominators. Cancel a factor in a numerator for a common, factor in a denominator. Multiply numerators and. denominators. Answer: 1/3 Note that unlike signs yield a negative product. Thus, (15/8) ■ (—14/9) -35/12. □ 1.3. THE RATIONAL NUMBERS 35 Dividing Fractions Every nonzero rational number has was it called a multiplicative inverse or reciprocal. The Reciprocal. If a is any nonzero rational number, then l/a is called the multiplicative inverse or reciprocal of a, and: 1 a- - = 1 a Note that: and 3 5 5 ' 3 and 1. Thus, the reciprocal of 2 is 1/2, the reciprocal of 3/5 is 5/3, and the reciprocal of —4/7 is —7/4. Note that to find the reciprocal of a number, simply invert the number (flip it upside down). Now we can define the quotient of two fractions. Division of Fractions. If a/b and c/d are two fractions, then their quotient is defined as follows: a _ c ad b d b c That is, dividing by c/d is the same as multiplying by the reciprocal d/c. The above definition of division is summarized by the phrase "invert and mul- tiply." You Try It! 35 / EXAMPLE 5. Simplify: -=- ( - 1CA 12/ Simplif 4 27 v: '-. Q 81 Solution: Invert and multiply, then factor in place and cancel common factors in a numerator for common factors in a denominator. 35 . / 10\ 35 / 12\ ~2i ~ v 12/ ~~ ~2i ' v Toy Invert and multiply. 5-7 ( 2-2-3" 3 • 7 V 2 • 5 , I Prime factor. $•1 ( t-1-P t-1 \ %•$ , 1 Cancel common factors. 2 ~ 1 Multiply numerators and denominators. = 2 Simplify. 36 CHAPTER 1. THE ARITHMETIC OF NUMBERS Answer: -4/3 Note that when all the factors in a denominator cancel, a 1 remains. Thus, (—35/21) -T- (—10/12) = 2. Note also that like signs yield a positive result. □ Adding Fractions First the definition. Addition of Fractions. If two fractions have a denominator in common, add the numerators and place the result over the common denominator. In symbols: a b a + b For example: 3 7 _ 4 ~5 + 5 ~ 5 and 11 T and If the fractions do not posses a common denominator, first create equivalent fractions with a least common denominator, then add according to the rule above. Least Common Denominator. If the fractions a/b and c/d do not share a common denominator, the least common denominator for b and d, written LCD(&,eQ, is defined as the smallest number divisible by both b and d. You Try It! Simplify: 1 — 1 y 6 9 EXAMPLE 6. Simplify: 5_ 12 Solution: The least common denominator in this case is the smallest number divisible by both 8 and 12. In this case, LCD(8, 12) = 24. We first need to make equivalent fractions with a common denominator of 24. Answer: -13/18 8 + 12 3 3 5 ~8 ' 3 + 12 2 2 9 10 ~24 + 24 1 24 Make equivalent fraction with a common denominator of 24. Multiply numerators and denominators. Add: -9 + 10 = 1. Note how we add the numerators in the last step, placing the result over the common denominator. Thus, —3/8 + 5/12 = 1/24. □ 1.3. THE RATIONAL NUMBERS 37 Order of Operations Rational numbers obey the same Rules Guiding Order of Operations as do the integers. Rules Guiding Order of Operations. When evaluating expressions, pro- ceed in the following order. 1. Evaluate expressions contained in grouping symbols first. If grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols first. 2. Evaluate all exponents that appear in the expression. 3. Perform all multiplications and divisions in the order that they appear in the expression, moving left to right. 4. Perform all additions and subtractions in the order that they appear in the expression, moving left to right. You Try It! EXAMPLE 7. Given x = 2/3, y = -3/5, and z = 10/9, evaluate xy + yz. Solution: Following Tips for Evaluating Algebraic Expressions, first replace all occurrences of variables in the expression xy + yz with open parentheses. Next, substitute the given values of variables (2/3 for x, —3/5 for y, and 10/9 for z) in the open parentheses. Given a = -1/2, b= 2/3, and c = —3/4, evaluate the expression a + be and simplify the result. xy + yz 10 Replace variables with parentheses. Substitute: 2/3 for x, —3/5 for y, and 10/9 for z. 38 CHAPTER 1. THE ARITHMETIC OF NUMBERS Answer: Use the Rules Guiding Order of Operations to simplify. 6 / 30\ ~~15 + V 45/ Multiply. -H-l) Reduce. 2 3 / 2 5\ ~~5 ' 3 + V 3 ' 5/ Make equival 6 / 10\ = -15 + ( v -15J least common 16 ~ ~15 Add. Thus, if x = 2/3, y = —3/5, and z = 10/9, then xy + yz = —16/15 □ You Try It! Simplify: -1/3) 4 EXAMPLE 8. Given x = -3/5, evaluate -x 3 . Solution: First, replace each occurrence of the variable x with open paren- theses, then substitute —3/5 for x. Answer: 1/81 ( r 3 ~5 3 5 27 125 27 125 Replace x with open parentheses. Substitute —3/5 for x. Write —3/5 as a factor three times. The product of three negative fractions is negative. Multiply numerators and denominators. The opposite of -27/125 is 27/125. Hence, 27/125, given x = -3/5. □ You Try It! Given x = —3/4 and y = —4/5, evaluate x 2 — y 2 . EXAMPLE 9. Given a = -4/3 and b = -3/2, evaluate a 2 + 2ab - 3b 2 . Solution: Following Tips for Evaluating Algebraic Expressions, first replace all occurrences of variables in the expression a 2 + lab— 3& 2 with open parentheses. 1.3. THE RATIONAL NUMBERS 39 Next, substitute the given values of variables (—4/3 for a and —3/2 for b) in the open parentheses. a 2 + 2ab - 36 2 Next, evaluate the exponents: (-4/3) 2 = 16/9 and (-3/2) 2 = 9/4. 16 2 3 /9 i u Next, perform the multiplications and reduce. _ 16 24 27 ~ ~9 + ~6 ~ T 16 27 = h4 9 4 Make equivalent fractions with a common denominator, then add. 16 4 36 27 9 ~ T ' 4 + ' 36 ~ T ' 9 _ 64 144 243 ~ 36 + "36" ~ lifT 35 ~~36 Thus, if a = -4/3 and b = -3/2, then a 2 + 2al 3b 2 -35/36 Answer: -31/400 □ Fractions on the Graphing Calculator We must always remember that the graphing calculator is an "approximating machine." In a small number of situations, it is capable of giving an exact answer, but for most calculations, the best we can hope for is an approximate answer. However, the calculator gives accurate results for operations involving frac- tions, as long as we don't use fractions with denominators that are too large for the calculator to respond with an exact answer. 40 CHAPTER 1. THE ARITHMETIC OF NUMBERS You Try It! Simplify using the graphing calculator: 4 8 "5 + 3 EXAMPLE 10. Use the graphing calculator to simplify each of the following expressions: 1 2 Solution: We enter each expression in turn. (a) - V ; 3 y ' 3 7 M 3 . 1 (C) 5^3 a) The Rules Guiding Order of Operations tell us that we must perform divi- sions before additions. Thus, the expression 2/3 + 1/2 is equivalent to: 2/3 2 1 1/2= - + - 1 3 2 4 3 ~ 6 + 6 _ 7 ~ 6 Divide first. Equivalent fractions with LCD. Add. Enter the expression 2/3+1/2 on your calculator, then press the ENTER key. The result is shown in the first image in Figure 1.12. Next, press the MATH button, then select l:^Frac (see the second image in Figure 1.12) and press the ENTER key again. Note that the result shown in the third image in Figure 1.12 matches the correct answer of 7/6 found above. 2/3+1/2 BJJ1I1 HUM CPtt PRE 2/3+1/2 1.166666667 ija^Frac 1.166666667 I 2: ►Dec 3:^ 4:*J-< 5: x j 6ifMin<: 7-1-fMaxC flns^Frac 7/6 I Figure 1.12: Calculating 2/3+1/2. b) The Rules Guiding Order of Operations tell us that there is no preference for division over multiplication, or vice-versa. We must perform divisions and multiplications as they occur, moving from left to right. Hence: /3x 5/7 = 2xV7 2 Divide: 2/3 = - ^ 2 10 Multiply: - x 5 = — 10 1 T x 7 Invert and multiply. 10 21 10 1 10 Multiply: — x - = — y - y 3 7 21 1.3. THE RATIONAL NUMBERS 41 This is precisely the same result we get when we perform the following calculation. 10 21 Hence: 2/3x5/7 is equivalent to 2 5 - x - 3 7 Multiply numerators and denominators. 2 5 - x - 3 7 Enter the expression 2/3x5/7 on your calculator, then press the ENTER key. The result is shown in the first image in Figure 1.13. Next, press the MATH button, then select 1 : ►Frac (see the second image in Figure 1.13) and press the ENTER key again. Note that the result shown in the third image in Figure 1.13 matches the correct answer of 10/21 found above. 2/3*5/7 araii hum cpx pre 2/3*5/7 .4761904762 Ija^Frac .4761904762 I 2: ►Dec 3:^ 4:*J-< 5: x j fcifMinC 7-lfMaxC fins ► Frac 10/21 Figure 1.13: Calculating 2/3 x 1/2. This example demonstrates that we need a constant reminder of the Rules Guiding Order of Operations. We know we need to invert and multiply in this situation. 3 3 - x - 5 1 9 5 Invert and multiply. Multiply numerators and denominators. So, the correct answer is 9/5. Enter the expression 3/5/1/3 on your calculator, then press the ENTER key. Select l:^Frac from the MATH menu and press the ENTER key again. Note that the result in the first image in Figure 1.14 does not match the correct answer of 9/5 found above. What have we done wrong? If we follow the Rules Guiding Order of Operations exactly, then: 3/5/1/3 =g/l/3 -> _ 3 1 ~ 5 X 3 1 Divide: 3/5 = 3 Divide: -/l 3 5 3 ' 5 Invert and multiply. 3 11 Multiply: - x - = - 5 3 5 42 CHAPTER 1. THE ARITHMETIC OF NUMBERS Answer: 28/15 This explains the answer found in the first image in Figure 1.14. However, it also show that: 3 ; 1 5^3 3/5/1/3 is not equivalent to We can cure the problem by using grouping symbols. Parentheses first. / is equivalent to -j- u o Hence: (3/5)/(l/3)= |/| _ 3 _ 1 ~ 5 ~ 3 (3/5)/(l/3) is equivalent to 3 , 1 5^3 Enter the expression (3/5)/ (1/3) on your calculator, then press the EN- TER key. Select 1 : ►Frac from the MATH menu and press the ENTER key again. Note that the result in the second image in Figure 1.14 matches the correct answer of 9/5. 3/5/1/3 fins ► Frac .2 1/5 (3/5)/(l/3) fins ► Frac l.S 9/5 Figure 1.14: Calculating (3/5)/(l/3). □ 1.3. THE RATIONAL NUMBERS 43 **.**.**. Exercises -*s -« ■*? In Exercises 1-6, reduce the given fraction to lowest terms by dividing numerator and denominator by the their greatest common divisor. 20 36 1. — 4. — 50 14 2. H 5* 38 45 10 21 3. — 6. — 48 36 In Exercises 7-13, reduce the given fraction to lowest terms by prime factoring both numerator and denominator and cenceling common factors. ~ 153 ,~ 171 7. 10. 170 144 198 159 8. 11. 144 106 188 140 9. 12. 141 133 In Exercises 13-18, for each of the following problems, multiply numerators and denominators, then prime factor and cancel to reduce your answer to lowest terms. 13 18/2 14. — ■ — 16 V 5 15.-H.f_i? 4 V 13 16. 3 ~2 ' H? 17. 16 ~~8~ ' 19 18. 14 7 17 In Exercises 19-24, for each of the following problems, first prime factor all numerators and denomina- tors, then cancel. After canceling, multiply numerators and denominators. 5 / 12\ 36 / 21 19. 20. 6 V 49/ 17 V 46 44 CHAPTER 1. THE ARITHMETIC OF NUMBERS 21. 22. 21 12 10 ' 55 49 52 13 ' 51 23. 55 29 ' (-s 24. 7 13 ' / 55 ^~49 In Exercises 25-30, divide. Be sure your answer is reduced to lowest terms. 75 50 / 5 39 V 58 76. 25 {-!) 77. 60 . ~17^ 34 "31 ?8 27 . 45 28 23 79. 7 10 { 28 30. 4 ~13 ~ / 48 " V 35 In Exercises 31-38, add or subtract the fractions, as indicated, and simplify your result. 31.4 37. 33. — 9 34. 35 "4 36. — 2 37. 9 5 4 1 38. 7 3 In Exercises 39-52, simplify the expression. 39. 40. 41. 5 2 2 ~ 5 7 1 6 ~ 2 2 «■ 1 1 1 2 iV 5 44. 45. 46. 47. 4MB 43. 48. 1.3. THE RATIONAL NUMBERS 45 49. 50. 6 2 5 5 3 5 (i 51. 52. In Exercises 53-70, evaluate the expression at the given values. 53. xy z at x -1/2, y = -1/3, and z = 5/2 54. xy—z 2 at x = —1/3, y = 5/6, and z = 1/3 55. -5a; 2 + 2y 2 at x = 3/4 and y = -1/2. 56. -2.x 2 + Ay 2 at x = 4/3 and y = -3/2. 57. 2a; 2 - 2xy - 3y 2 at x = 3/2 and y = -3/4. 58. 5a; 2 - 4xy- 3y 2 at a; = 1/5 and y = -4/3. 59. x + yz at x = —1/3, y = 1/6, and z = 2/5. 60. x + yz a,t x = 1/2, y = 7/4, and 2 = 2/3. 61. ab+bc at a = -4/7, 6 = 7/5, and c = -5/2 62. ab+bc at a = -8/5, 6 = 7/2, and c = -9/7 63. x 3 at a; = -1/2 64. a; 2 at a; = -3/2 65. x—yz at a; = —8/5, y = 1/3, and z = —8/5 66. x — yz at x = 2/3, y = 2/9, and z = —3/5 67. -a; 2 at a; = -8/3 68. -x 4 at x = -9/7 69. a; 2 + yz at a; = 7/2, y = -5/4, and z = -5/3 70. a; 2 + yz at x = 1/2, y = 7/8, and z = -5/9 71. a + b/c + d is equivalent to which of the following mathematical expressions? a + b (a) a H h d c (c) + d (b) c + d (d)a- c c + d 72. (a + 6)/c + d is equivalent to which of the following mathematical expressions? a + b (a) a H h d c , . a + 6 (c) + d (b) c + d (d)o- c + d 73. a + b/(c + d) is equivalent to which of the following mathematical expressions? , + b (a) a H h d c (c) + d (b) c + d (d)o- c c + d 74. (a + 6)/(c + d) is equivalent to which of the following mathematical expressions? a + b (a) a H h d c , . a + 6 (c) + d (d) a + - 75. Use the graphing calculator to reduce 4125/1155 to lowest terms. 76. Use the graphing calculator to reduce 2100/945 to lowest terms. 40 CHAPTER 1. THE ARITHMETIC OF NUMBERS 77. Use the graphing calculator to simplify 45 70 84 ' 33' 78. Use the graphing calculator to simplify 34 13 55 + 77' 79. Use the graphing calculator to simplify 28 33 35 44 80. Use the graphing calculator to simplify 11 84 11 30 f». f». 5». Answers ■*i ■*; -m 2 1. - 5 3. A 24 15 9 To 4 9 -3 11. 3 - 2 13. 15. 45 13 171 "20" 10 19. — 49 21. 23. 120 '275 270 "29" 25. 580 ~"39" 27. 930 ~289 29. 98 05 31. 7 ~12 33. 11 ~~9 35. 1 ~30 37. 70 ~45 39. 109 ~"90" 41. 79 30 43. 53 35 45. 47. 131 "1T 323 "^0" 02 49. — 45 1.3. THE RATIONAL NUMBERS 47 5 16 51. 65. 42 15 53 _73 67. -64/9 12 43 37 69. — 55.-— 3 16 57. «! 16 71. (a) 73. (d) 2 59 - "g 75. 25/7 43 77. 25/22 ol. 10 79. 16/15 63. -1/8 48 CHAPTER 1. THE ARITHMETIC OF NUMBERS You Try It! Change 24/7 to a decimal. 1.4 Decimal Notation Every rational number can be expressed using decimal notation. To change a fraction into its decimal equivalent, divide the numerator of the fraction by its denominator. In some cases the process will terminate, leaving a zero remain- der. However, in other cases, the remainders will begin to repeat, providing a decimal representation that repeats itself in blocks. EXAMPLE 1. , , 39 Change each of the following fractions into decimals. v ; 11 Solution: We perform two divisions, the one on the left to change 39/80 to a decimal, the one on the right to find a decimal representation for 4/11. 0.3636 0.4875 80)39.0000 32 7 00 6 40 600 560 400 400 1)4.0000 33 70 66 40 33 70 66 4 On the left, the division process terminates with a zero remainder. Hence, 39/80 = 0.4875 is called a terminating decimal. On the right, the remainders repeat in a pattern and the quotient also repeats in blocks of two. Hence, 4/11 = 0.3636 ... is called a repeating decimal. We can also use a repeating bar to write 4/11 = 0.36. The block under the repeating bar repeats itself Answer: 3.428571 indefinitely. □ You Try It! Change 0.45 to a fraction. Reduce to lowest terms. Vice-versa, any terminating decimal can be expressed as a fraction. You need only count the number of digits after the decimal point and use the same number of zeros in your denominator. EXAMPLE 2. Express each of the following decimals as fractions. Reduce your answers to lowest terms. (a) 0.055 (b) 3.36 Solution: In each case, count the number of digits after the decimal point and include an equal number of zeros in the denominator. 1.4. DECIMAL NOTATION 49 In example (a), there are three dig- In example (b), there are two digits its after the decimal point, so we after the decimal point, so we place place the number over 1000, which the number over 100, which has two has three zeros after the one. zeros after the one. 55 336 0.055 = 3.36 = 1000 100 11 _ 84 ~ 200 ~ 25 Answer: 9/20 □ As we saw in Example 1, the repeating decimal 0.36 is equivalent to the fraction 4/11. Indeed, any repeating decimal can be written as a fraction. For example, 0.3 = 1/3 and 0.142857 = 1/7. In future courses you will learn a technique for changing any repeating decimal into an equivalent fraction. However, not all decimals terminate or repeat. For example, consider the decimal 0.42422422242222..., which neither terminates nor repeats. This number cannot be expressed using repeating bar notation because each iteration generates one additional 2. Be- cause this number neither repeats nor terminates, it cannot be expressed as a fraction. Hence, 0.42422422242222... is an example of an irrational number. Irrational numbers. If a number cannot be expressed in the form p/q, where p and q are integers, q ^ 0, then the number is called an irrational number. Real numbers. By including all of the rational and irrational numbers in one set, we form what is known as the set of real numbers. The set of real numbers includes every single number we will use in this text- book and course. Adding and Subtracting Decimals When adding signed decimals, use the same rules you learned to use when adding signed integers or fractions. Sign rules for addition. When adding two decimal numbers, use the follow- ing rules: • To add two decimals with like signs, add their magnitudes and prefix their common sign. 50 CHAPTER 1. THE ARITHMETIC OF NUMBERS • To add two decimals with unlike signs, subtract the smaller magnitude from the larger, then prefix the sign of the decimal number having the larger magnitude. You Try It! Simplify: -22.6+18.47 Answer: -4.13 EXAMPLE 3. Simplify: (a) -2.3 + (-0.015) and (b) 6.95 Solution: In part (a), note that we have like signs. Hence, we add the mag- nitudes and prefix the common sign. 2.300 +0.015 -2.3 + (-0.015) = -2.315 2315~ In part (b), note that we have unlike signs. Thus, we first subtract the smaller magnitude from the larger magnitude, then prefix the sign of the decimal num- ber with the larger magnitude. 8.40 -6.95 6.95 -1.45 1.45 Hence, -2.3 + (-0.015) = -2.315 and -8.4 + 6.95 = -1.45. □ You Try It! Simplify: -22.6 - 18.47 Answer: -41.07 Subtraction still means "add the opposite." EXAMPLE 4. Simplify: (a) -5.6 - 8.4 and (b) -7.9 - (-5.32) Solution: In part (a), first we add the oppposite. Then we note that we have like signs. Hence, we add the magnitudes and prefix the common sign. 5.6 -5.6-8.4 = -5.6 +(-8.4) +8.4 = -14.0 14-0 In part (b), first we add the opposite. Then we note that we have unlike signs. Thus, we first subtract the smaller magnitude from the larger magnitude, then prefix the sign of the decimal number with the larger magnitude. 7.90 -7.9- (-5.32) = -7.9 + 5.32 -5.32 = -2.58 2.58 Hence, —5.6 -14.0 and -7.9- (-5.32) -2.58. □ 1.4. DECIMAL NOTATION 51 Multiplication and Division of Decimals The sign rules for decimal multiplication and division are the same as the sign rules used for integers and fractions. Sign Rules for multiplication and division. When multiplying or dividing two decimal numbers, use the following rules: • Like signs give a positive result. • Unlike signs give a negative result. Multiplication of decimal numbers is fairly straightforward. First multiply the magnitudes of the numbers, ignoring the decimal points, then count the number of digits to the right of the decimal point in each factor. Place the decimal point in the product so that the number of digits to the right of the decimal points equals the sum of number of digits to the right of the decimal point in each factor. EXAMPLE 5. Simplify: (-1.96)(2.8) Solution: Multiply the magnitudes. The first decimal number has two digits to the right of the decimal point, the second has one digit to the right of the decimal point. Thus, we must place a total of three digits to the right of the decimal point in the product. 1.96 x2.8 1 568 3 92 5.488 (-1.96)(2.8) = -5.488 Note that unlike signs yield a negative product. You Try It! Simplify: (-12.5)(-23.4) Answer: 292.50 When dividing signed decimal numbers, ignore the signs and divide the magnitudes. Push the decimal point in the divisor to the end of the divisor, then move the decimal point in the dividend an equal number of spaces. This sets the decimal point in the quotient. □ You Try It! EXAMPLE 6. Simplify: -4.392 -r (-0.36) Solution. Divide the magnitudes. Move the decimal in the divisor to the end of the divisor. Move the decimal in the dividend an equal number of places (two places) to the right. Simplify: -5.76/3.2 52 CHAPTER 1. THE ARITHMETIC OF NUMBERS 0.36 4.39 2 Answer: Place the decimal point in the quotient directly above the new position of the decimal point in the dividend, then divide. 12.2 36)439.2 36 79 72 72 72 Like signs yield a positive result. Hence, —4.392-=- (—0.36) = 12.2. □ Order of Operations Decimal numbers obey the same Rules Guiding Order of Operations as do the integers and fractions. Rules Guiding Order of Operations. When evaluating expressions, pro- ceed in the following order. 1. Evaluate expressions contained in grouping symbols first. If grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols first. 2. Evaluate all exponents that appear in the expression. 3. Perform all multiplications and divisions in the order that they appear in the expression, moving left to right. 4. Perform all additions and subtractions in the order that they appear in the expression, moving left to right. You Try It! Given y = —0.2, evaluate: EXAMPLE 7. Given x = -0.12, evaluate -x 2 . Solution: Following Tips for Evaluating Algebraic Expressions, first replace all occurrences of variable x in the expression — x 2 with open parentheses. Next, 1.4. DECIMAL NOTATION 53 substitute —0.12 for x in the open parentheses, then simplify. \2 -X -(-0.12) z -(0.0144) -0.0144 Replace x with open parentheses. Substitute —0.12 for x. Exponent: (-0.12) 2 = 0.0144 Negate. Note that we square first, then we negate second. Thus, if x = —0.12, then -x 2 = -0.0144. Answer: -0.0016 □ EXAMPLE 8. Given -0.3, evaluate 1.2a: 2 — 3 Ax. Solution: Following Tips for Evaluating Algebraic Expressions, first replace all occurrences of variable x in the expression 1.2a; 2 — 3.4a; with open parentheses. Next, substitute —0.3 for x in the open parentheses, then simplify. You Try It! Given y = —0.15, evaluate: -lAy 2 + 2.2y 1.2a; 2 - 3.4a; 1.2( ) 2 -3.4( ) 1.2(-0.3) 2 -3.4(-0.3) 1.2(0.09) -3.4(-0.3) 0.108- (-1.02) 0.108+1.02 1.128 Replace x with parentheses. Substitute —0.3 for x. Exponent: (-0.3) 2 = 0.09. Multiply: 1.2(0.09) = 0.108 and 3.4(-0.3) = -1.02. Add the opposite. Simplify. Thus, if x = -0.3, then 1.2a; 2 - 3.4a; = 1.128. Answer: -0.3615 □ We saw earlier that we can change a fraction to a decimal by dividing. You Try It! EXAMPLE 9. Given x = 2/5, evaluate -3.2.x + 5. Given y = -3/4, evaluate: Solution: Following Tips for Evaluating Algebraic Expressions, first replace _« o . 7 all occurrences of variable x in the expression —3.2a; + 5 with open parentheses. Next, substitute 2/5 for x in the open parentheses. -3.2a; + 5 = -3.2 Replace x with open parentheses. -3.2 ( - ) + 5 Substitute 2/5 for x. 54 CHAPTER 1. THE ARITHMETIC OF NUMBERS Answer: 8.725 You Try It! Given z = —0.4, evaluate: 4 5- -z 5 One approach is to change 2/5 to a decimal by dividing the numerator by the denominator. Thus, 2/5 = 0.4. = -3.2(0.4) + 5 Replace 2/5 with 0.4. = -1.28 + 5 Multiply: -3.2(0.4) = -1.28. = 3.72 Add: -1.28+5 = 3.72. Thus, if x = 2/5, then -3.2a; + 5 = 3.72. □ As we saw in Example 2, we can easily change a terminating decimal into a fraction by placing the number (without the decimal point) over the proper power of ten. The choice of the power of ten should match the number of digits to the right of the decimal point. For example: 411 , _ 311 151111 100 0.411 and 3.11 and 15.1111 1000 100 10000 Note that the number of zeros in each denominator matches the number of digits to the right of the decimal point. EXAMPLE 10. Given y -0.25, evaluate y + 4. Solution: Following Tips for Evaluating Algebraic Expressions, first replace all occurrences of variable y in the expression — (3/5)y + 4 with open parentheses. Next, substitute —0.25 for y in the open parentheses. :2/ + 4 (-0.25) Replace y with open parentheses. Substitute —0.25 for y. Answer: 133/25 Place 25 over 100 to determine that —0.25 -0.25 = -1/4. -25/100, or after reduction, 1 20 3 80 20 + 20 83 20 + 4 Replace -0.25 with -1/4. Multiply: -| (-1) = A. Make equivalent fractions with LCD. Add. Thus, if y = -0.25, then -(3/5)j/ + 4 = 83/20. □ 1.4. DECIMAL NOTATION 55 Rounding Using the Graphing Calculator Here is the algorithm for rounding a decimal number to a particular place. Rules for rounding. To round a number to a particular place, follow these steps: 1. Mark the place you wish to round to. The digit in this place is called the rounding digit. 2. Mark the digit in the place to the immediate right of the rounding digit. This is called the test digit. a) If the test digit is greater than or equal to 5, add 1 to the rounding digit, then replace all digits to the right of the rounding digit with zeros. Trailing zeros to the right of the decimal point may be deleted. b) If the test digit is less than 5, keep the rounding digit the same, then replace all digits to the right of the rounding digit with zeros. Trailing zeros to the right of the decimal point may be deleted. EXAMPLE 11. Use your graphing calculator to evaluate 125a; 3 — 17.5a;+44. at x = —3.13. Round your answer to the nearest tenth. Solution. First, store —3.13 in the variable X with the following keystrokes. 0QQ0 X,T,e,n I ENTER The result is shown in the first image in Figure 1.15. Next, enter the expression 125a; 3 — 17. 5x + 44.8 with the following keystrokes. ED LUSH X,T,0,n ENTER The result is shown in the second image in Figure 1.15. Thus, the answer is approximately —3733.462125. We now need to round this answer to the nearest tenth. Mark the rounding digit in the tenths place and the test digit to its immediate right. Test digit Rounding digit 3733. |T| [t\ 2125 A You Try It! Evaluate a; 3 — 3a; at x = -1.012. Round to the nearest hundredth. 56 CHAPTER 1. THE ARITHMETIC OF NUMBERS 125*X A 3-17.5*X+4 4.S -3733.462125 Figure 1.15: Evaluate 125a; 3 - 17.5a; + 44.8 at x = -3.13. Because the test digit is greater than or equal to 5, add 1 to the rounding digit, then replace all digits to the right of the rounding digit with zeros. -3733.462125: -3733.500000 Answer: 2.0 Delete the trailing zeros from end of the fractional part of a decimal. This does not change our answer's value. -3733.462125 « -3733.5 Therefore, if x = —3.13. then to the nearest tenth: 125a; 3 - 17.5a; + 44.6 -3733.5 □ 1.4. DECIMAL NOTATION 57 ti. ;». ;». Exercises ■*j ■*-: •*; In Exercises 1-33, simplify the given expression. 1. -2.835 +(-8.759) 2. -5.2 +(-2) 3. 19.5- (-1.6) 4. 9.174- (-7.7) 5. -2-0.49 6. -50.86-9 7. (-1.2)(-0.05) 8. (-7.9)(0.9) 9. -0.13 + 23.49 10. -30.82 + 75.93 11. 16.4 +(-41.205) 12. -7.8 + 3.5 13. -0.4508 + 0.49 14. 0.2378 +(-0.29) 15. (-1.42)(-3.6) 16. (-8.64) (4.6) 17. 2.184+ (-0.24) 18. 7.395 + (-0.87) 19. (-7.1)(-4.9) 20. (5.8)(-1.9) 21. 7.41 + (-9.5) 22. -1.911 + 4.9 23. -24.08 + 2.8 24. 61.42 +(-8.3) 25. (-4.04)(-0.6) 26. (-5.43)(0.09) 27. -7.2- (-7) 28. -2.761- (-1.5) 29. (46.9)(-0.1) 30. (-98.9) (-0.01) 31. (86.6)(-1.9) 32. (-20.5)(8.1) In Exercises 33-60, simplify the given expression. 33. -4.3- (-6.1)(-2.74) 34. -1.4- 1.9(3.36) 35. -3.49+ |-6.9- (— 15.7)| 36. 1.3 + | - 13.22-8.79| 37. |18.9 - 1.55| - | - 16.1 - (-17.04) | 38. | - 17.5 - 16.4| - | - 15.58 - (-4.5) | 39. 8.2- (-3.1) 3 40. -8.4- (-6.8) 3 41. 5.7- (-8.6)(1.1) 2 42. 4.8-6.3(6.4) 2 43. (5.67)(6.8)- (1.8) 2 44. (-8.7)(8.3)- (-1.7) 2 45. 9.6 +(-10.05- 13.16) 46. -4.2 + (17.1- 14.46) 47. 8.1 + 3.7(5.77) 48. 8.1 + 2.3(-5.53) 58 CHAPTER 1. THE ARITHMETIC OF NUMBERS 49. 7.5 + 34.5/(-1.6 + 8.5) 50. -8.8 + 0.3/(-7.2 + 7.3) 51. (8.0 + 2.2)/5.1 - 4.6 52. (35.3 + 1.8)/5.3- 5.4 53. -18.24- | - 18.5- 19. 7| 54. 16.8- 14.58- 17.141 55. -4.37- | -8.97| 56. 4.1- | -8.4| 57. 7.06- (-1.1-4.41) 58. 7.74- (0.9- 7.37) 59. -2.2- (-4.5) 2 60. -2.8- (-4.3) 2 61. Evaluate a — b 2 at a = —2.9 and b = —5.4. 62. Evaluate a — b 3 at a = —8.3 and b = —6.9. 63. Evaluate a+\b-c\ at a = -19.55, b = 5.62, and c = —5.21. 64. Evaluate a — \b — c\ at a -8.31, and c = 17.5. .37, 65. Evaluate a — be at a = 4.3, b = 8.5, and c= 1.73. 66. Evaluate a + be at a = 4.1, b = 3.1, and c= -7.03. 67. Evaluate a — (b — c) at a = —7.36, b = -17.6, and c= -19.07. 68. Evaluate \a — b\ — \c — d\ at a = 1.91, b = 19.41, c = -11.13, and d = 4.3. 69. Evaluate a + b/(c + d) at a = 4.7, b = 54.4, c = 1.7, and d = 5.1. 70. Evaluate (a + b)/c - d at a = -74.2, 6 = 3.8, c = 8.8, andd= 7.5. 71. Evaluate ab — c 2 at a = —2.45, b = 5.6, and c = —3.2. 72. Evaluate a + (b - c) at a = 12.6, b = -13.42, andc= -15.09. 73. Evaluate a- \b\ at a = -4.9 and b = -2.67. 74. Evaluate a - be 2 at a = -3.32, b = -5.4, and c = —8.5. 75. Use your graphing calculator to evaluate 3.5 — 1.7x at x = 1.25 Round your answer to the nearest tenth. 76. Use your graphing calculator to evaluate 2.35a: — 1.7 at x = —12.23 Round your answer to the nearest tenth. 77. Use your graphing calculator to evaluate 1.7a; — 3.2x + 4.5 at x = 2.86 Round your answer to the nearest hundredth. 78. Use your graphing calculator to evaluate 19.5 — 4.4a; — 1.2a; 2 at x = —1.23 Round your answer to the nearest hundredth. 79. Use your graphing calculator to evaluate —18.6 + 4.4a: 2 — 3.2a; 3 at x = 1.27 Round your answer to the nearest thousandth. 80. Use your graphing calculator to evaluate —4.4a; 3 — 7.2a; — 18.2 at x = 2.29 Round your answer to the nearest thousandth. 1.4. DECIMAL NOTATION 59 s*- j*- £*■ Answers <#$ >#s <#$ 1. -11.594 41. 16.106 3. 21.1 43. 35.316 5. -2.49 45. -13.61 7. 0.06 47. 29.449 9. 23.36 49. 12.5 11. -24.805 51. -2.6 13. -0.92 53. -56.44 15. 5.112 55. -13.34 17. -9.1 57. 12.57 19. 34.79 59. -22.45 21. -0.78 61. -32.06 23. -8.6 63. -8.72 25. 2.424 65. -10.405 27. -0.2 67. -8.83 29. -4.69 69. 12.7 31. -164.54 71. -23.96 33. -21.014 73. -7.57 35. 5.31 75. 1.4 37. 16.41 77. 9.25 39. 37.991 79. -4.948 c,0 CHAPTER 1. THE ARITHMETIC OF NUMBERS You Try It! Simplify: 2(3x) Answer: 6x 1.5 Algebraic Expressions The associative property of multiplication is valid for all numbers. Associative Property of Multiplication. Let a, b, and c be any numbers. Then: a ■ (b ■ c) = (a ■ b) ■ c The associative property of multiplication is useful in a number of situations. EXAMPLE 1. Simplify: -3 (Ay) Solution: Currently the grouping — 3(Ay) demands that we first multiply 4 and y. However, we can use the associative property of multiplication to regroup, first multiplying —3 and 4. -3(Ay) = (-3 ■ A)y = -12y Thus, -3(4y) = -12y. The associative property of multiplication. Multiply: -3-4= -12 □ Let's look at another example. You Try It! Simplify: -3(-8u 2 Answer: 2Au 2 EXAMPLE 2. Simplify: -2(-Axy) Solution: Currently, the grouping — 2(— Axy) demands that we first multiply —4 and xy. However, we can use the associative property of multiplication to regroup, first multiplying —2 and —4. — 2(— Axy) = (—2 • (—A))xy The associative property of multiplication. = %xy Multiply: -2 ■ (-4) = 8 Thus, — 2(— Axy) = 8xy. □ In practice, we can move quicker if we perform the regrouping mentally, then simply write down the answer. For example: -2(-4t) = 8t and 2(-5z 2 ) = -10z 2 and - 3(4w 3 I2u j 1.5. ALGEBRAIC EXPRESSIONS 61 The Distributive Property We now discuss a property that couples addition and multiplication. Consider the expression 2 • (3 + 5). The Rules Guiding Order of Operations require that we first simplify the expression inside the parentheses. 2 ■ (3 + 5) = 2 • 8 = 16 Add: 3 + 5 = 8 Multiply: 2 • 8 16 Alternatively, we can instead distribute the 2 times each term in the parenthe- ses. That is, we will first multiply the 3 by 2, then multiply the 5 by 2. Then we add the results. 2- (3 + 5) = 2-3 + 2-5 = 6 + 10 = 16 Distribute the 2. Multiply: 2 • 3 = 6 and 2 • 5 = 10 Add: 6+10 = 16 Note that both methods produce the same result, namely 16. This example demonstrates an extremely important property of numbers called the distribu- tive property. The Distributive Property. Let a, b, and c be any numbers. Then: a- (b + c) = a b + a ■ c That is, multiplication is distributive with respect to addition. EXAMPLE 3. Use the distributive property to expand 2(3x + 7). Solution: First distribute the 2 times each term in the parentheses. Then simplify. 2(3x + 7) = 2(3x) + 2(7) Use the distributive property. = 6a; + 14 Multiply: 2 (3a;) = 6x and 2(7) = 14 Thus, 2(3x + 7) = 6x + 14. You Try It! Expand: 5(2y+ 7) Answer: Wy + 35 □ Multiplication is also distributive with respect to subtraction. You Try It! EXAMPLE 4. Use the distributive property to expand — 2(5?/ — 6). Expand: -3(2z-7) <i2 CHAPTER 1. THE ARITHMETIC OF NUMBERS Solution: Change to addition by adding the opposite, then apply the dis- tributive property. -2(5y - 6) = -2(5y + (-6)) Add the opposite. = — 2(5y) + (— 2)(— 6) Use the distributive property. = -Wy+ 12 Multiply: -2(5y) = -Wy and (-2)(-6) = 12 Answer: -6z + 21 Thus, -2(5y - 6) = -Wy + 12. □ Some might prefer to "take their time," using the technique shown in Examples 3 and 4 until they feel "ready" to move more quickly. Speeding Things Up a Bit In Example 4, we changed the subtraction to addition, applied the distributive property, then several steps later we were finished. However, if you understand that subtraction is really the same as adding the opposite, and if you are willing to do a few steps in your head, you should be able to simply write down the answer immediately following the given problem. If you look at the expression — 2(5y — 6) from Example 4 again, only this time think "multiply —2 times by, then multiply —2 times —6, then the result is immediate. -2(5y-6) = -10y+12 Let's try this "speeding it up" technique in a couple more examples. You Try It! -2x - 12) Expand: — 3(— 2a + 36—7) EXAMPLE 5. Use the distributive property to expand — 3(- Solution: To distribute the —3, we simply think as follows: "— 3(— 2x) = 6x, — 3(5y) = — 15y, and — 3(— 12) = 36." This sort of thinking allows us to write down the answer immediately without any additional steps. -5y- -2x) Answer: 6a — 96 + 21 -3(-2x + 5y- 12) = 6a; - 15y + 36 □ You Try It! Expand: -4(— x - 2y - 7) Answer: 4a; + 8y + 28 EXAMPLE 6. Use the distributive property to expand — 5(— 2a — 56 + 8). Solution: To distribute the —5, we simply think as follows: "— 5(— 2a) = 10a, — 5(— 56) = 256, and —5(8) = —40." This sort of thinking allows us to write down the answer immediately without any additional steps. -5(-2a-56- 10a + 256 -40 □ 1.5. ALGEBRAIC EXPRESSIONS 63 Distributing a Negative Sign Recall that negating a number is equivalent to multiplying the number by — 1 . Multiplicative Property of Minus One. If a is any number, then: (— l)a = —a This means that if we negate an expression, it is equivalent to multiplying the expression by —1. You Try It! EXAMPLE 7. Expand -{7x- 8y - 10). Expand: -(-a -26 +11) Solution: First, negating is equivalent to multiplying by —1. Then we can change subtraction to addition by "adding the opposite" and use the distribu- tive property to finish the expansion. — (7x — Sy — 10) = — l(7a; — 8y — 10) Negating is equivalent to multiplying by — 1 = -l(7x + (-8y) + (-10)) Add the opposite. = -l(7x) + (-l)(-8y) + (-1)(-10) Distribute the -1. = -7x + 8y + 10 Multiply. Thus, -(7x -8y- 10) = -7x + 8y + 10. Answer: a + 26 - 11 □ While being mathematically precise, the technique of Example 7 can be simpli- fied by noting that negating an expression surrounded by parentheses simply changes the sign of each term inside the parentheses to the opposite sign. Once we understand this, we can simply "distribute the minus sign" and write: -(7a; -8y- 10) = -7x + 8y + 10 In similar fashion, — (—3a + 56 — c) = 3a — 56 + c and -(-3a; - 8y + 11) = 3x + 8y- 11. 04 CHAPTER 1. THE ARITHMETIC OF NUMBERS You Try It! Simplify: 3y + 8y Answer: lly Combining Like Terms We can use the distributive property to distribute a number times a sum. a(b + c) = ab + ac However, the distributive property can also be used in reverse, to "unmultiply" or factor an expression. Thus, we can start with the expression ab + ac and "factor out" the common factor a as follows: ab + ac = a(b + c) You can also factor out the common factor on the right. ac + be = (a + b)c We can use this latter technique to combine like terms. EXAMPLE 8. Simplify: 7x + 5x Solution: Use the distributive property to factor out the common factor x from each term, then simplify the result. 7a; + 5a; = (7 + 5)a; = 12x Thus, 7a; + 5a; = 12a;. Factor out an x using the distributive property. Simplify: 7 + 5 = 12 □ You Try It! Simplify: — 5z + 9z 3 _!_ o~3 EXAMPLE 9. Simplify: -8a 2 + 5a 2 Solution: Use the distributive property to factor out the common factor o 2 from each term, then simplify the result. Answer: 4^ la 2 + 5a 2 = (-8 + 5)a 2 -3a 2 Factor out an a using the distributive property. Simplify: —8 + 5 = —3 Thus, -8a 2 + 5a 2 -3a 2 a Examples 8 and 9 combine what are known as "like terms." Examples 8 and 9 also suggest a possible shortcut for combining like terms. 1.5. ALGEBRAIC EXPRESSIONS 65 Like Terms. Two terms are called like terms if they have identical variable parts, which means that the terms must contain the same variables raised to the same exponents. For example, 2x 2 y and llx 2 y are like terms because they contain identical variables raised to the same exponents. On the other hand, — 3si 2 and As 2 t are not like terms. They contain the same variables, but the variables are not raised to the same exponents. Consider the like terms 2x 2 y and llx 2 y. The numbers 2 and 11 are called the coefficients of the like terms. We can use the distributive property to combine these like terms as we did in Examples 8 and 9, factoring out the common factor x 2 y. 2x 2 y + llx 2 y= (2 + ll)x 2 y = 13x 2 y However, a much quicker approach is simply to add the coefficients of the like terms, keeping the same variable part. That is, 2 + 11 = 13, so: 2x 2 y + llx 2 y 13x 2 y This is the procedure we will follow from now on. EXAMPLE 10. Simplify: -8w 2 + 17w 2 Solution: These are like terms. If we add the coefficients —8 and 17, we get 9. Thus: -8w 2 + 17w 2 9w 2 Add the coefficients and repeat the variable part. You Try It! Simplify: Aab — \bab Answer: — llafc a EXAMPLE 11. Simplify: -Auv - 9uv Solution: These are like terms. If we add —4 and —9, we get —13. Thus: — Auv — 9uv = —I3uv Add the coefficients and repeat the variable part. You Try It! Simplify: —3xy — 8xy Answer: — llxy a You Try It! EXAMPLE 12. Simplify: -3x 2 y + 2xy 2 Simplify: 5ab + llbc (i(» CHAPTER 1. THE ARITHMETIC OF NUMBERS Answer: 5ab + llbc You Try It! Simplify: -3z 2 + Az- 8z 2 -9z Solution: These are not like terms. They do not have the same variable parts. They do have the same variables, but the variables are not raised to the same exponents. Consequently, this expression is already simplified as much as possible. -3x 2 y + 2xy 2 Unlike terms. Already simplified. □ Sometimes we have more than just a single pair of like terms. In that case, we want to group together the like terms and combine them. EXAMPLE 13. Simplify: -8u - Av - 12m + 9v Solution: Use the associative and commutative property of addition to change the order and regroup, then combine line terms. — 8m — Av — 12m + 9v = (—8m — 12m) + (—Av + 9v) Reorder and regroup. = — 20m + 5m Combine like terms. Answer: — 11 z 2 — 5z Note that -8m- 12m -20m and — Av + 9v = 5m. Alternate solution. You may skip the reordering and regrouping step if you wish, simply combining like terms mentally. That is, it is entirely possible to order your work as follows: -8m- Av- 12m + 9m -20m + 5m Combine like terms. □ In Example 13, the "Alternate solution" allows us to move more quickly and will be the technique we follow from here on, grouping and combining terms mentally. Order of Operations Now that we know how to combine like terms, let's tackle some more compli- cated expressions that require the Rules Guiding Order of Operations. Rules Guiding Order of Operations. When evaluatin g expressions, pro- ceed in the following order. 1. Evaluate expressions contained in grouping symbols first. If grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols first. 1.5. ALGEBRAIC EXPRESSIONS 67 2. Evaluate all exponents that appear in the expression. 3. Perform all multiplications and divisions in the order that they appear in the expression, moving left to right. 4. Perform all additions and subtractions in the order that they appear in the expression, moving left to right. EXAMPLE 14. Simplify: 4(-3a + 26) - 3(4a - 56) Solution: Use the distributive property to distribute the 4 and the —3, then combine like terms. 4(-3a + 26) - 3(4a - 56) = - 12a + 86 - 12a + 156 Distribute. = —24a + 236 Combine like terms. Note that -12a- 12a: -24a and 156 = 236. You Try It! Simplify: -2x - 3(5 - 2cc) Answer: 4a; — 15 □ You Try It! EXAMPLE 15. Simplify: -2(3x - Ay) - (5a; - 2y) Solution: Use the distributive property to multiply —2 times 3a; — Ay, then distribute the minus sign times each term of the expression 5a;— 2y. After that, combine like terms. -2(3cc - Ay) - (5a: - 2y) = -6x + 8y- 5x + 2y Distribute. = — 11a; + 10y Combine like terms: Note that —6a; — 5a; = —lis and 8y + 2y = lOy. Simplify: -3(u + v) — (u — 5v) Answer: — Au + 2v □ EXAMPLE 16. Simplify: -2{x 2 y - 3a;y 2 ) - A(-x 2 y + ixy 2 ) Solution: Use the distributive property to multiply —2 times x 2 y — 3xy 2 and —4 times —x 2 y + 3a;y 2 . After that, combine like terms. -2{x 2 y - 3xy 2 ) - A(-x 2 y + 3xy 2 ) = -2x 2 y + 6xy 2 + Ax 2 y - 12xy 2 = 2x y — 6xy Note that —2x 2 y + Ax 2 y = 2x 2 y and 6a;y 2 — 12a;y 2 = —6xy 2 . You Try It! Simplify: 8u 2 v - 3(u 2 v + Auv 2 ) Answer: 5u 2 v — 12uv 2 D G8 CHAPTER 1. THE ARITHMETIC OF NUMBERS You Try It! Simplify: x - 2[-x + 4(x + 1)} When grouping symbols are nested, evaluate the expression inside the in- nermost pair of grouping symbols first. EXAMPLE 17. Simplify: -2x - 2 (-2x - 2 [-2x - 2]) Solution: Inside the parentheses, we have the expression — 2x — 2[— 2a; — 2]. The Rules Guiding Order of Operations dictate that we should multiply first, expanding —2 [—2x — 2] and combining like terms. -2x - 2 (-2.T - 2 [-2x - 2]) = -2x - 2 (-2x + Ax + 2) = -2x- 2 (2a; + 2) In the remaining expression, we again multiply first, expanding — 2(2a; + 2) and combining like terms. Answer: — 5x -2a; - 4a; - 4 -6a; -4 □ 1.5. ALGEBRAIC EXPRESSIONS 69 **.**.**. Exercises ■*$*$-*$ In Exercises 1-6, use the associative property of multiplication to simplify the expression. Note: You must show the regrouping step using the associative property on your homewwork. 1. -3(6o) 4. 8{5xy) 2. -W(2y) 5. -7(3a; 2 ) 3. -9(6o6) 6. -6(8z) In Exercises 7-18, use the distributive property to expand the given expression. 7.1(3x-7y) 13. -(-3u - 6v + 8) 8. -4(5a + 26) 14. -(3u - 3v - 9) 9. -6(-y + 9) 15. -8(4u 2 - 6v 2 ) 10. 5(-9u> + 6) 16. -5 (8a; - 9y) 11. -9(s + 9) 17. -(7u + 10^ + 8) 12. 6(-10y + 3) 18. -(7u-8v-5) In Exercises 19-26, combine like terms by first using the distributive property to factor out the common variable part, and then simplifying. Note: You must show the factoring step on your homework. 19. -19a; + 17a; - 17a; 23. 9y 2 x + Uy 2 x - 3y 2 x 20. lira- 3n- 18n 24. 4a; 3 - 8a; 3 + 16a; 3 21. 14a; 3 - 10a; 3 25. 15rra + Um 22. -lly 3 -6y 3 26. 19 q + 5q In Exercises 27-38, simplify each of the following expressions by rearranging and combining like terms mentally. Note: This means write down the problem, then write down the answer. No work. 27. 9- 17m- m + 7 31. -5m - 16 + 5 - 20m 28. -11 + 20a; + 16a; - 14 32. -18q + 12 - 8 - 19 q 29. -6y 2 - 3a; 3 + Ay 2 + 3a; 3 33. -16x 2 y + 7y 3 - 12y 3 - 12x 2 y 30. Uy 3 - lly 2 x + lly 3 + 10y 2 x 34. 10a: 3 + Ay 3 - 13y 3 - 14a; 3 70 CHAPTER 1. THE ARITHMETIC OF NUMBERS 35. -Ur + 16 - 7r - 17 37. 14 - 16w - 10 - 13m 36. -9s- 5- 10s + 15 38. 18 + lOcc + 3 - 18a; In Exercises 39-58, use the distributive property to expand the expression, then combine like terms mentally. 39. 3 - (-5m + 1) 49. -7- (5 + 3x) 40. 5- (-10? + 3) 50. 10 -(6 -4m) 41. -(9y 2 + 2x 2 ) - 8(5y 2 - 6a; 2 ) 51. -8(-5m - 8) - 7(-2 + Qy) 42. -8{-8y 2 + Ax 3 ) - 7(3m 2 + x 3 ) 52. 6(-3s + 7) - (4 - 2s) 43. 2(10 - 6p) + 10(-2p + 5) 53. 4(-7m 2 - 9x 2 y) - 6(-5x 2 y - by 2 ) 44. 2(3 - 7x) + (-7x + 9) 54. -6(a; 3 + 3y 2 x) + 8(-y 2 x - 9x 3 ) 45. 4(-10n + 5) - 7(7n - 9) 55. 6s - 7 - (2 - 4s) 46. 3(-9n + 10) + 6(-7n + 8) 56. 4.x - 9 - (-6 + 5a;) 47. -Ax - 4 - (10a; - 5) 57. 9(9 - lOr) + (-8 - 2r) 48. 8y + 9 - (-8y + 8) 58. -7(6 + 2p) + 5(5 - hp) In Exercises 59-64, use the distributive property to simplify the given expression. 59. -7a; + 7(2a; - 5[8ir + 5]) 62. 2x + A(5x - 7[8a; + 9]) 60. -9a; + 2(5a; + 6 [-8a; - 3]) 63. -8a; - 5(2a; - 3[-4a; + 9]) 61. 6a; - 4(-3a; + 2[5a; - 7]) 64. 8a; + 6(3a; + 7[-9a; + 5]) s*> j*> j*' Answers ■*& •*& •** 1. -18a 11. -9s -81 3. -54a6 13. 3u + 6v - 8 5. -21a; 2 15. -32m 2 + 48w 2 7. 12a;-28w 17. -7m - 10m - 8 9. 6m -54 19. -19a; 1.5. ALGEBRAIC EXPRESSIONS 71 21. 4a; 3 43. 70 - 32p 23. 19y 2 x 45. -89n + 83 25.29m 47. -14a; +1 27. 16- 18m 49. -12 -3a; 29. -2y 2 51. -23y + 78 31. -25m -11 53. 2y 2 - 6x 2 y 33. -28x 2 y - 5y 3 55. 10s - 9 35. -21r-l 57. 73 - 92r 37. 4-29y 59. -273a; - 175 39. 2 + 5y 61. -22a; + 56 41. -49y 2 + 46a; 2 63. -78a; + 135 Chapter 2 Solving Linear Equations and Inequalities In this chapter, we will begin solving linear equations by using inverse opera- tions to solve for the unknown variable. Humans have been solving problems of this type for thousands of years. The earliest records indicate that the problems were written entirely in words with no symbols (such as +, — , x, -=-, and =) or numbers. For example, a problem might read "Sixteen added to an unknown number is fifty-two." Diophantes of Alexandria wrote a series of books Arithmetica demonstrat- ing solutions to equations in which he used some abbreviations for shortcuts. In 9th century AD, al-Khwarizmi wrote the book Al-Kitab al-mukhtasar ti Hisab al-jabr w'al-muqabala, translated as The Compendious Book on Calcu- lation by Restoration and Reduction. In fact, the word "algebra" is based on the word al-Jabr from the text title. His book described the rules of "Comple- tion and Reduction" to make it easier to do arithmetic computations needed in human business interactions. In the 1600-1700's, mathematicians began writing out equations symboli- cally as we do today using the symbols +, — , x, and -j- for operations and = for equals. We will be solving equations which can be solved in logical steps by using inverse operations which "undo" each other. For example, one pair of inverse operations is addition and subtraction. Another pair of inverse operations is multiplication and division. We will also be solving inequalities which may have an infinite set of answers that can be expressed in three ways: by graphing on a number line, by using set-builder notation, or by using interval notation. 73 74 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 2.1 Solving Equations: One Step Let's start with the definition of an equation. Equation. An equation is a mathematical statement that equates two alge- braic expressions. The key difference between a algebraic expression and an equation is the pres- ence of an an equals sign. So, for example, 2x + 3, x - (3 - 2x), and 2(y + 3) - 3(1 - y) are algebraic expressions, while 2a; + 3 = 0, x - (3 - 2x) = 4, and 2{y + 3) - 3(1 - y) = -11 are equations. Note that each of the equations contain an equals sign, but the algebraic expressions do not. Next we have the definition of a solution of an equation. What it Means to be a Solution of an Equation. A solution of an equation is a numerical value that satisfies the equation. That is, when the variable in the equation is replaced by the solution, a true statement results. You Try It! Which of the numbers {1, 2, 3, 4, 5} is a solution of the equation 2y + 3 = 7. Answer: EXAMPLE 1. Show that 8 is a solution of the equation x — 12 = —4. Solution. Substitute 8 for x in the given equation and simplify. 12 = -4 12 = -4 -4 = -4 The given equation. Substitute 8 for x. Simplify both sides. Since the left- and right-hand sides of the last line are equal, this shows that when 8 is substituted for x in the equation a true statement results. Therefore, 8 is a solution of the equation. □ Equivalent Equations Now that we know how to identify a solution of an equation, let's define what it means when we say that two equations are equivalent. 2.1. SOLVING EQUATIONS: ONE STEP 75 Equivalent Equations. Two equations are equivalent if they have the same solution set. You Try It! EXAMPLE 2. Are the equations x — 3 = 6 and x = 9 equivalent? Solution. The number 9 is the only solution of the equation x — 3 = 6 Similarly, 9 is the only solution of the equation x = 9. Therefore x — 3 = 6 and x = 9 have the same solution sets and are equivalent. Answer Are the equations x = 5 and X — 7 = 10 equivalent? No. □ You Try It! EXAMPLE 3. Are the equations x 2 = 1 and x = 1 equivalent? Solution. By inspection, the equation x 2 = 1 has two solutions, —1 and 1. Are the equations x = 1 and x 2 = x equivalent? (-1) 2 1 and 1 1 On the other hand, the equation x = 1 has a single solution, namely 1. Hence, the equations x 2 = 1 and x = 1 do not have the same solution sets and are not equivalent. Answer: No. As we shall soon see, equivalent equations play an important role in finding the solution of an equation. □ Wrap and Unwrap, Do and Undo Suppose that you are wrapping a gift for your cousin. You perform the following steps in order. 1. Put the gift paper on. 2. Put the tape on. 3. Put the decorative bow on. When we give the wrapped gift to our cousin, he politely unwraps the present, "undoing" each of our three steps in inverse order. 1. Take off the decorative bow. 2. Take off the tape. 3. Take off the gift paper. 76 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES This seemingly frivolous wrapping and unwrapping of a gift contains some deeply powerful mathematical ideas. Consider the mathematical expression x + 4. To evaluate this expression at a particular value of x, we would start with the given value of x, then add 4. • Let's set x equal to the number 7. If we add 4, we arrive at the following result: 11 Now, how would we "unwrap" this result to return to our original number? We would start with our result, namely 11, then subtract 4. • Take our result from above, 11. If we subtract 4, we return to our original value of x: 7 The above discussion leads us to two extremely important observations. The inverse of addition is subtraction. If we start with a number x and add a number a, then subtracting a from the result will return us to the original number x. In symbols, x + a — a = x. That is, subtracting a "undoes" the effect of adding a and returns us to the original number x. The inverse of subtraction is addition. If we start with a number x and subtract a number a, then adding a to the result will return us to the original number x. In symbols, x — a + a = x. That is, adding a "undoes" the effect of subtracting a and returns us to the original number x. Operations that Produce Equivalent Equations In Example 1, we saw that x = 8 was a solution of the equation x — 12 = —4. Indeed, the equations x = 8 and x — 12 = —4 are equivalent equations because both have the same solution sets. In the case of the equation x — 12 = —4, it's fairly easy to "guess" that the solution is x = 8, but as the equations become more complicated, we will want to know just what operations we can perform on the equation that will not change the solution set. The goal is to start with an equation such as a; -12 = -4, then through a series of steps that do not change the solution, arrive at the solution 2.1. SOLVING EQUATIONS: ONE STEP 77 With these thoughts in mind, there are a number of operations that will pro- duce equivalent equations (equations with the same solution set). The first two that we will employ are adding or subtracting the same amount from both sides of an equation. Adding the Same Quantity to Both Sides of an Equation. Adding the same quantity to both sides of an equation does not change the solution set. That is, if a = b, then adding c to both sides of the equation produces the equivalent equation a + c = b + c. Subtracting the Same Quantity from Both Sides of an Equation. Subtracting the same quantity to both sides of an equation does not change the solution set. That is, if a = b, then subtracting c from both sides of the equation produces the equivalent equation a — c = b — c. Let's look at an example where adding the same amount to both sides of the equation produces an equivalent equation that is the solution to the original equation. You Try It! EXAMPLE 4. Solve x - 7 = 12 for x. Solve for ,r: Solution: To undo the effect of subtracting 7, we add 7 to both sides of the x — 6 equation. x — 7 = 12 Original equation. x — 7 + 7=12 + 7 Adding 7 to both sides of the equation produces an equivalent equation. x = 19 On the left, adding 7 "undoes" the effect of subtracting 7 and returns x. On the right, 12 + 7= 19. Therefore, the solution of the equation is 19. 78 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Answer: 10 You Try It! Solve for x: 1 _ 3 2 ~~ 5 Check: To check, substitute the solution 19 into the original equation. x-7=12 19-7 = 12 12 = 12 Original equation. Substitute 19 for x. Simplify both sides. The fact that the last line of the check is a true statement guarantees that 19 is a solution of x — 7 = 12. □ In the solution of Example 4, we use the concept of the "inverse." If we start with x, subtract 7, then add 7, we are returned to the number x. In symbols, x — 7 + 7 = x. We are returned to x because "subtracting 7" and "adding 7" are inverse operations of one another. That is, whatever one does, the other "undoes." Next, let's look at an example where subtracting the same amount from both sides of the equation produces an equivalent equation that is the solution to the original equation. EXAMPLE 5. Solve for Solution: To undo the effect of adding 2/3, we subtract 2/3 from both sides of the equation. 2 1 x + - = - 3 2 _ 3 4 X ~6~6 Original equation. Subtracting 2/3 from both sides produces an equivalent equation. On the left, subtracting 2/3 "undoes" the effect of adding 2/3 and returns x. On the right, make equivalent fractions with a common denominator. 34 ! Subtract: = 6 6 6 Therefore, the solution of the equation is —1/6. Check: Let's use the TI-84 graphing calculator to check this solution. 1. Store the value —1/6 in the variable X using the following keystrokes. The result is shown in Figure 2.1. EDQBH ENTER 2.1. SOLVING EQUATIONS: ONE STEP 79 2. Enter the left-hand side of the original equation: x + 2/3. Use the fol- lowing keystrokes. The result is shown in Figure 2.1. X,T,0,n B0B0 -l/6-»X 1666666667 Figure 2.1: Checking the solution to x + 2/3 = 1/2. 3. Press the MATH button on your calculator (see Figure 2.2), then select l:^Frac and press the ENTER button. This will convert the decimal result to a fraction (see Figure 2.2). BJJ1I1 HUN CPtt PRE -l/6-»X ija^Frac -. 1666666667 2: ►Dec X+2/3 3:* .5 4:*J-< flns^Frac 5: *f 1/2 6:fMin<: I 7-i-fMaxC Figure 2.2: Changing the result to a fraction. 1 1 2 1 Note that the result is — , showing that is a solution of x H — = — . 2 6 3 2 Answer: 1/10 □ More Operations That Produce Equivalent Equations Here are two more operations that produce equivalent equations. Multiplying Both Sides of an Equation by a Nonzero Quantity. Mul- tiplying both sides of an equation by a nonzero quantity does not change the solution set. That is, if a = b, and c / 0, then multiplying both sides of the equation by c produces the equivalent equation ac = be. 80 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Dividing Both Sides of an Equation by a Nonzero Quantity. Dividing both sides of an equation by a nonzero quantity does not change the solution set. That is, if a = b, and c/0, then dividing both sides of the equation by c produces the equivalent equation a b Like addition and subtraction, multiplication and division are inverse opera- tions. The inverse of multiplication is division. If we start with a number x and multiply by a number a, then dividing the result by the same number a returns us to the original number x. In symbols, That is, dividing by a "undoes" the effect of multiplying by a and returns us to the original number x. The inverse of division is multiplication. If we start with a number x and divide by a number a, then multiplying the result by the same number a returns us to the original number x. In symbols, That is, multiplying by a "undoes" the effect of dividing by a and returns us to the original number x. Let's look at an example where dividing both sides of the equation by the same amount produces an equivalent equation that is the solution to the original equation. You Try It! Solve for x: -3.6a; = 0.072 EXAMPLE 6. Solve -2.1a; = 0.42 for Solution: To undo the effect of multiplying by —2.1, we divide both sides of 2.1. SOLVING EQUATIONS: ONE STEP si -2.1a; = 0.42 -2.1a; 0.42 -2.1 -2.1 x = -2 the equation by —2.1. Original equation. Dividing both sides by —2.1 produces an equivalent equation. On the left, dividing by —2.1 "undoes" the effect of multiplying by —2.1 and returns x. On the right, divide: 0.42/(-2.1) = -2. Therefore, the solution of the equation is —2. Check: To check, substitute the solution —2 into the original equation. — 2.1a; = 0.42 Original equation. -2.1(-2) = 0.42 Substitute -2 for x. 0.42 = 0.42 On the left, multiply: -2.1(-2) = 0.42 The fact that the last line of the check is a true statement guarantees that is a solution of —2.1a; = 0.42. Answer: -0.02 □ Next, lets look at an example where multiplying both sides of the equation by the same amount produces an equivalent equation that is the solution to the original equation. EXAMPLE 7. Solve -10 for Solution: To undo the effect of dividing by 5, we multiply both sides of the equation by 5. L5 X 5 ~~ -10 ; J = [-10] 5 x = -50 Original equation. Multiplying both sides by 5 produces an equivalent equation. On the left, multiplying by 5 "undoes" the effect of dividing by 5 and returns x. On the right, multiply: [— 10]5 = —50. Therefore, the solution of the equation is —50. Check: Let's use the TI-84 graphing calculator to check this solution. 1. Store the value —50 in the variable X using the following keystrokes. The result is shown in Figure 2.3. Eomm ENTER You Try It! Solve for x: x 7 82 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 2. Enter the left-hand side of the original equation: — . Use the following 5 keystrokes. X,T,0,n E The result is shown in Figure 2.3. -50-»X X/5 -50 -10 Figure 2.3: Checking the solution to x/5 = —10. Answer: —14 Note that the result is —10, showing that —50 is a solution of x/5 = —10. □ Writing Mathematics When solving equations, observe the following rules when presenting your work. 1. One equation per line. This means that you s hould not arrange your work horizontally as follows: x + 3 = 7 x + 3 -3 = 7 -3 x = 4 That s three equations on a line. Instead, work vertically, writing one equat ion per line. In the following presentation, note how we align the equal signs in a column. x + 3 = 7 z + 3- 3 = 7- x = 4 3 2.1. SOLVING EQUATIONS: ONE STEP 83 2. Add and subtract inline. Do not add 7 to both sides of the equation in the following manner: x-7 = + 7 x = : 12 +7 = 19 Instead, add 7 ' 'inline" to both sides of the equation as follows: x-7 = = 12 X -7+7 = x = = 12 + 7 = 19 84 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES ;*■;-*■ ti. Exercises •*;■*; ■*■; In Exercises 1-6, which of the numbers following the given equation are solutions of the given equation? Support your response with work similar to that shown in Example 1. 1. x + 2 = 4 2. £ + 6 = 9 3. £-9 = 4 3, 2, 9, 5 4, 6, 3, 10 16, 14, 20, 13 4.x- 3 = 5; 9,15,8,11 5. a;-3 = 6; 9,16,10,12 6. x-5 = 7; 15,13,19,12 In Exercises 7-12, are the following equations equivalent? 7. £ — 1 = —7 and x = —8 8. £ — 7 = —8 and x = —15 9. £ — 5 = —5 and x = 10. £ — 3 = —3 and x = 11. £ 2 = 1 and £ = 1 12. £ 2 = 16 and x = 4 In Exercises 13-32, solve the given equation for x. 13. £-20 = 9 14. £- 10 = 5 15. 16 = £-3 16. 16 = £-8 17. £ + 11 = 20 18. £ + 10= 18 19. 9 = 2-19 20. 4 = £- 11 21. 20 = 9 + £ 22. 18 = 7 + £ 23. 18= 17 + £ 24. 15 = 7 + £ 25. 7 + £ = 19 26. 16 + £ = 17 27. £ - 9 = 7 28. £-2 = 8 29. £ + 15 = 19 30. £ + 6 = 10 31. 10 + £ = 15 32. 18 + £ = 19 In Exercises 33-40, solve the equation and simplify your answer. 4 2 33. £ = - 9 7 1 1 34. £ - - = - 3 2 2.1. SOLVING EQUATIONS: ONE STEP 85 7 4 11 35.* + - = -- 38 - + I = 3 19 9 1 36. x + - = -- 39.x = — 2 7 8 2 5 7 5 2 37. #+- = - 40. a; = — 9 2 9 3 In Exercises 41-46, solve the given equation for x. 41. -5.1a; = -12.75 42. —3.5a; = -22.4 43. -6.9x = -58.65 44. -1.4a; = -4.34 45. -3.6a; = -24.12 46. -6.4a; = -39.68 In Exercises 47-52, solve the given equation for x. 47. - = -11 2 48. - = 12 49. - = -18 50. 1~" 51. ^ = * 52. l = -7 j»- s*/ 5*> Answers >** •** •** 1. 2 17. 9 3. 13 19 - 28 21. 11 23. 1 25. 12 27. 16 11. No 29. 4 5. 9 7. No 9. Yes 13. 29 31 - 5 46 15. 19 33 - 63 86 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 43. 8.5 45. 6.7 47. -22 49. -144 41.2.5 51.-105 35. 79 ~36 37. 53 18 39. 5 2.2. SOLVING EQUATIONS: MULTIPLE STEPS 87 2.2 Solving Equations: Multiple Steps Recall the "Wrap" and "Unwrap" discussion from Section 2.1. To wrap a present we put the gift paper on, put the tape on, and put the decorative bow on. To unwrap the gift, we must "undo" each of these steps in inverse order. Hence, to unwrap the gift we take off the decorative bow, take off the tape, and take off the gift paper. Now, imagine a machine that multiplies its input by 3, then adds 5 to the result. This machine is pictured on the left in Figure 2.4. 1. Multiply by 3. 2. Add 5. 1. Subtract 5. 2. Divide by 3. Figure 2.4: The second machine "unwraps" the first machine. To "unwrap" the effect of the machine on the left, we need a machine that will "undo" each of the steps of the first machine, but in inverse order. The "unwrapping" machine is pictured on the right in Figure 2.4. It will first sub- tract 5 from its input, then divide the result by 3. Note that each of these operations "undoes" the corresponding operation of the first machine, but in inverse order. The following argument shows that the second machine "undoes" the op- eration of the first machine. 1. Drop the integer 4 into the machine on the left in Figure 2.4. This ma- chine will first multiply 4 by 3, then add 5 to the result. The result is 3(4) + 5, or 17. 2. To "unwrap" this result, drop 17 into the machine on the right. This machine first subtracts 5, then divides the result by 3. The result is (17 — 5)/3, or 4, the original integer that was put into the first machine. EXAMPLE 1. Solve for x: 3x + 5 = 14 You Try It! Solve for Solution: On the left, order of operations demands that we first multiply x by 3, then add 5. To solve this equation for x, we must "undo" each of these operations in inverse order. Thus, we will first subtract 5 from both sides of 2x + 3 = 7 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES the equation, then divide both sides by 3. 3a; + 5 = 14 3a; + 5-5 = 14 -5 3a; = 9 3a: _ 9 T ~ 3 x = 3 Original equation. To "undo" adding 5, subtract 5 from both sides of the equation. Simplify both sides. To "undo" multiplying by 3, divide both sides of the equation by 3. Simplify both sides. Check: To check the solution, substitute 3 for x in the original equation and simplify. 3a; + 5 = 14 3(3) + 5= 14 9 + 5= 14 14= 14 Original equation. Substitute 3 for x. Multiply first: 3(3) = 9. Add: 9 + 5 = 14. Answer: x = 2 Because the last line of the check is a true statement, this guarantees that 3 is a solution of the original equation. □ You Try It! Solve for x: x 3 _ 1 2 ~ 5 ~ 4 Let's try an equation with fractions. EXAMPLE 2. Solve for x 2 _ 1 5 ~ 3 ~ 2 Solution: On the left, order of operations demands that we first divide x by 5, then subtract 2/3. To solve this equation for x, we must "undo" each of these operations in inverse order. Thus, we will first add 2/3 to both sides of the equation, then multiply both sides of the resulting equation by 5. x 2 _ 1 5 ~ 3 ~ 2 x 2 2 _ 1 2 Original equation. To "undo" subtracting 2/3, add 2/3 to both sides of the equation. On the left, we simplify. On the right, we make equivalent fractions with a common denominator. x 5 ~ x _ 7 5 ~ 6 3 4 6 + 6 Make equivalent fractions 3 4 7 Add: 6 6 6' 2.2. SOLVING EQUATIONS: MULTIPLE STEPS s<) Now we "undo" dividing by five by multiplying both sides of the equation by 5. 5 (I) 35 Multiply both sides by 5. On the left, simplify. On the right, multiply: 5 ( -J = —. Check: Let's use the TI-84 to check this solution. 1. Store the value 35/6 in the variable X using the following keystrokes. HEBE ENTER The result is shown in Figure 2.5. 2. Enter the left-hand side of the original equation: x/5 — 2/3. Use the following keystrokes. EBEBE ENTER The result is shown in Figure 2.5). 5.833333333 .5 3. Press the MATH button on your calculator (see Figure 2.6), then select l:^Frac, then press the ENTER button. This will convert the decimal Figure 2.5: Checking the so- result to a fraction (see Figure 2.6). lution to x/5 - 2/3 = 1/2. BjaU HUM CPtt PRE 35/6-»X ija^Frac 5.S33333333 Z: ►Dec y./5-2/Z 3:^ .5 4:*J-< flns^Frac 5: *f 1/2 6:fMin<: I 7-i-fMaxC Figure 2.6: Changing the result to a fraction. 1 35 x Note that the result is — , showing that — is a solution of — 2 s 6 5 Answer: x = 17/10 □ Let's try an equation with decimals. 90 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES You Try It! Solve for x: 3.25- 1.2a; = 0.37 EXAMPLE 3. Solve for 5.2a; + 2.3 -3.94 Solution: On the left, order of operations demands that we first multiply x by 5.2, then add 2.3. To solve this equation for x, we must "undo" each of these operations in inverse order. Thus, we will first subtract 2.3 from both sides of the equation, then divide both sides by 5.2. Original equation. 2.3 To "undo" adding 2.3, subtract 2.3 from both sides. On the left, simplify. On the right, add: -3.94-2.3 = -6.24. To "undo" multiplying by 5.2, divide both sides by 5.2. On the left, simplify. On the right, divide: -6.24/5.2 = -1.2. 5.2a; + 2.3 = -3.94 5.2a; + 2.3 -2.3 = -3.94 5.2a; = -6.24 5.2a; -6.24 5.2 5.2 x = -1.2 Check: To check the solution, substitute —1.2 for x in the original equation and simplify. 5.2a; + 2.3 = -3.94 Original equation. 5.2(-1.2) + 2.3 = -3.94 Substitute —1.2 for x. -6.24 + 2.3 = -3.94 Multiply: 5.2(-1.2) = -6.24 -3.94 = -3.94 Add: -6.24 + 2.3 = -3.94. Answer: x = 2.4 Because the last line of the check is a true statement, this guarantees that —1.2 is a solution of the original equation. □ Variables on Both Sides of the Equation It is not uncommon that the variable you are solving for appears in terms on both sides of the equation. Consider, for example, the equation 2a; + 3 = 5 — 7a;. In cases like this, it is helpful to have a general understanding of what it means to "solve for a;." Solve for x. When asked to solve an equation for x, the goal is to manipulate the equation into the final form X = "Stuff", 2.2. SOLVING EQUATIONS: MULTIPLE STEPS 91 where "Stuff" is a valid mathematical expression that may contain other vari- ables, mathematical symbols, etc., but it must not contain any occurrence of the variable x. In this section, "Stuff" will always be a single number, but in Section 2.4, Formulae, "Stuff" will take on added complexity, including variables other than x. Strategy for solving for x. When asked to solve an equation for x, a common strategy is to isolate all terms containing the variable x on one side of the equation and move all terms not containing the variable x to the other side of the equation. Solution: We need to isolate all terms containing x on one side of the equation. We can eliminate 5a; from the right-hand side of 3 — 2x = 5a; + 9 by subtracting 5a; from both sides of the equation. 3 — 2x = 5a; + 9 Original equation. 3 — 2x — 5a; = 5a; + 9 — 5x Subtract 5a; from both sides. 3 — 7x = 9 Simplify both sides. Next, eliminate 3 from the left-hand side of the last equation by subtracting 3 from both sides of the equation. 3 — 7a; — 3 = 9 — 3 Subtract 3 from both sides. — 7a; = 6 Simplify both sides. Note how we have isolated all terms containing x on one side of the equation. Divide both sides by —7. Simplify both sides. -7x 6 -7 ^7 x = 6 ~7 You Try It! EXAMPLE 4. Solve 3 - 2a; = 5a; + 9 for x. Solve for 4a; + 7 = 5 - 8a; 92 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Answer: x = —1/6 Check: To check the solution, substitute —6/7 for x in the original equation. Original equation. Substitute —6/7 for x. Multiply: -2(-6/7) = 12/7 and 5(-6/7) = -30/7. Make equivalent fractions with a common denominator. 3-2x~- = 5a; + 9 2 B)^ -(-?)♦ 12 3 + y = 30 = -y + 9 21 12 y + y = 30 63 = -y + y Add. 33 _ 33 y ~ y Because the last line of the check is a true statement, this guarantees that —6/7 is a solution of the original equation. □ You Try It! Solve for x: 2x-(x-2) = 2(a; + 7) Simplifying Expressions When Solving Equations Sometimes we need to simplify expressions before we can isolate terms contain- ing x. EXAMPLE 5. Solve for x: 2(3x + 1) - 3(4 - 2x) = -34 Solution: We'll first simplify the expression on the left-hand side of the equation using the Rules Guiding Order of Operations. 2(3a; + l) -3(4 -2x) = -34 6a; + 2- 12 + 6a; = -34 12a;- 10 = -34 Original equation. Multiply: 2(3a; + 1) = 6x + 2. Multiply: -3(4 - 2a;) = -12 + 6x. Add: 6a; + 6a; = \2x. Add: 2- 12 = -10. To "undo" subtracting 10, we add 10 to both sides of the equation. 12a; - 10 + 10 = -34 + 10 Add 10 to both sides. 12a; = — 24 Simplify both sides. To "undo" multiplying by 12, we divide both sides by 12. -24 12a; x 12 -2 Divide both sides by 12. Simplify both sides. Check: Let's use the TI-84 to check this solution. 2.2. SOLVING EQUATIONS: MULTIPLE STEPS 93 1. First, store —2 in the variable X using the following keystrokes. EDS ENTER The result is shown in the first image in Figure 2.7. 2. Enter the left-hand side of original equation: 2(3o; + 1) — 3(4 — 2x). Use the following keystrokes. 2 D > □ HBH ENTER The result is shown in the second image in Figure 2.7. 2*<3*X+l)-3*<4-2 -34 Figure 2.7: Checking the solution to 2(3x + 1) - 3(4 - 2x) -34. Note that when —2 is substituted for x in the left-hand of the equation, the result is —34, equalling the right-hand side of the equation. Thus, the solution — 2 checks. Answer: x 12 □ You Try It! EXAMPLE 6. Solve for x: 2x - 5(3 - 2x) = A(x - 1) Solution: We'll first simplify the expressions on each side of the equation using the Rules Guiding Order of Operations. Solve for x: 5(1 -x) = 2(x + 3)- (x- 1) 2a; -5(3 -2a;) = i(x - 1) 2x- 15+ 10a; = 4x - 4 12a;- 15 = 4a; -4 Original equation. On the left, distribute the —5. On the right, distribute the 4. On the left, add: 2a; + lOx = 12a; 94 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Next, we need to isolate terms containing the variable x on one side of the equation. To remove the term 4a; from the right-hand side, we subtract Ax from both sides of the equation. \2x — 15 — 4a; = 4x — 4 — Ax Subtract 4a; from both sides. 8a; — 15 = —4 Simplify both sides. To remove the term —15 from the left-hand side, we add 15 to both sides of the equation. 8a; - 15 + 15 = -4 + 15 Add 15 to both sides. 8a; = 11 Simplify both sides. Finally, to "undo" multiplying by 8, we divide both sides by 8. — = — Divide both sides by 8. 8 8 y x = — Simplify both sides. 8 Check: Let's use the TI-84 to check this solution. 1. First, store 11/8 in the variable X using the following keystrokes. 1 X,T,0,n 1 I ENTER □ □HE The result is shown in the first image in Figure 2.8. 2. Enter the left-hand side of the original equation: 2x — 5(3 — 2a;). Use the following keystrokes. ENTER The result is shown in the second image in Figure 2.8. 3. Enter the right-hand side of the original equation: 4(a; — 1). Use the following keystrokes. The result is shown in the third image in Figure 2.8. |T| |] Q^ \T\ ENTER 2.2. SOLVING EQUATIONS: MULTIPLE STEPS m ll/8-»X 1.375 2*X-5*<3-2*X) 1.5 4*CX-1) 1.5 Figure 2.8: Checking the solution to 2x — 5(3 — 2x) = A(x — 1). There is no need to use the l:^Frac from the MATH menu this time. The fact that both sides of the equation evaluate to an identical 1.5 when x = 11/8 guarantees that 11/8 is a solution of 2x — 5(3 — 2x) = 4(x — 1). Answer: ■1/3 □ 96 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES t* ■ ;-*■ ;*■ Exercises ■*:>•*'">■*; In Exercises 1-16, solve the given equation for x. 1. 2a;-20 = -12 2. -Ax- 1 =3 3. -ll + 3a; = -44 4. -8+Ux = -22 5. -5a; + 17= 112 6. 3a; + 12 = 51 7. -16a;- 14 = 2 8. 4a; - 4 = 64 9.5- 13a; = 70 10. 11 + 10a; = 81 11. -2-9a; = -74 12. -4- 16a; = -100 13. 7- x = -7 14. 20 - 3a; = 35 15. -Ax + 14 = 74 16. -4a; +15 = 27 In Exercises 17-24, solve the equation and simplify your answer. 17. 18. 19. x 1 _ 9 7 ~ 3 ~ ~8 x 8 _ 4 8 ~ 3 ~ ~7 a; 4 _ 3 7 + 9 ~ 2 x 1 8 20. - + - = - 5 6 7 21. a; 2 4 2 + 3 ~ 7 22. a; 4 3 7 + 5 ~ 4 23. a; 9 5 5 ~ 2 ~ ~3 24. a; 8 3 5 ~ 9 ~ ~2 In Exercises 26-32, solve each equation. 25. 0.3a; + 1.7 = 3.05 26. -7.2a; + 2.9 = 64.10 27. 1.2a; + 5.2 = 14.92 28. -7.3a; +1.8 = -45.65 29. 3.5a; -3.7= -26.10 30. -1.4a; -4.7 = 5.80 31. -4.7a;- 7.4 = -48.29 32. -5.2a;- 7.2 = 38.04 In Exercises 33-44, solve each equation. 33. 13-9a; = 11 - 5a; 34. 11- 10a; = 13 - 4a; 2.2. SOLVING EQUATIONS: MULTIPLE STEPS 97 35. 11a; + 10= 19a; + 20 36. 20a; + 19 = 10a; + 13 37. 11- 15a; = 13- 19a; 38. 13- 11a; = 17 -5a; 39. 9a; + 8 = 4- 19a; 40. 10a; + 8 = 6 - 2a; 41. 7a; + 11 = 16- 18a; 42. llx + 8 = 2- 17a; 43. 12a; + 9 = 4a; + 7 44. 6a; + 3 = 16a; + 11 In Exercises 45-56, solve each equation. 45. 8(5x-3) -3(4a; + 6) =4 46. 6(3a;-8) - 6(4ai + 6) = 3 47. 2a; - 4(4 - 9a;) = 4(7a; + 8) 48. 4a; - 9(6 - 2x) = 2(5x + 7) 49. 2(6 - 2x) - (4a; - 9) = 9 50. 2(8 -5a;)- (2a; - 6) = 4 51. 3(5x-6)- 7(7a; + 9) = 3 52. 9(3a; - 7) - 9(2a; + 9) = 6 53. 2a; - 2(4 - 9a;) = 8(6a; + 2) 54. 3a; - 3(5 - 9a;) = 6(8a; + 2) 55. 2(7 -9a;)- (2a; - 8) = 7 56. 8(5 -2a;)- (8a; - 9) = 4 ;*•;-*■ i± Answers •*;■*■; ■*'> 1. 4 3. -11 5. -19 7. -1 9. -5 11. 8 13. 14 15. -15 17. 13S ~~24 133 21. 21 23. 85 25. 4.5 27. 8.1 29. -6.4 31. 8.7 1 33. 35. 37. 19. 18 98 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 39. 1 ~7 41. 1 5 43. 1 ~4 45. 23 14 47. 24 y 49. 3 2 51. 42 ~17 53. 6 ~7 55. 3 4 2.3. CLEARING FRACTIONS AND DECIMALS 99 2.3 Clearing Fractions and Decimals In this section we introduce techniques that clear fractions and decimals from equations, making the resulting equation a lot easier to solve. When clearing fractions from an equation, you will need to simplify products like the ones posed in the following examples. You Try It! EXAMPLE 1. Simplify: 12 (1 X) . ^^ Solution: When we multiply three numbers, such as 12, 2/3, and x, the /$ associative property of multiplication tells us that it does not matter which \ T x two numbers we multiply first. We use the associative property to regroup, then multiply numerators and denominators and simplify the result. (2 \ ( 2\ 12 —x 1 = I 12 • — 1 x Associative property of multiplication. 24 -x Multiply: 12 ■ 2 = 24. o 8x Divide: 24/3 = 8. Answer: 9x □ Example 1 shows all of the steps involved in arriving at the answer. How- ever, the goal in this section is to perform this calculation mentally. So we just "Multiply 12 and 2 to get 24, then divide 24 by 3 to get 8." This approach allows us to write down the answer without doing any work. 12 (-x) =8x You should practice this mental calculation until you can write down the answer without writing down any steps. You Try It! EXAMPLE 2. Simplify: 18 ( -x , Solution: This time we perform the calculations mentally. Multiply 18 and 2 /g to get 36, then divide 36 by 9 to get 4. 14 I —x 18 (§*) = Ax Answer: 6x □ 100 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES When the numbers get larger, the mental calculations become harder. For example, consider 72 , x In this case, the work "multiply 72 and 8 to get 576, then divide 576 by 9 to get 64" is a bit difficult to carry in your head. However, this is when the calculator comes to the rescue. You Try It! Use your calculator to simplify: l r EXAMPLE 3. Use your calculator to help simplify 72 ( -x Solution: Use your calculator to multiply 72 and 8, then divide by 9. Enter 72*8/9 and press the ENTER key. 72*3^9 64 Answer: 45a; Thus, 72 [ -x 1 = 64cc. □ Canceling is More Efficient In Examples 1, 2, and 3, we multiplied numerators, then divided by the sole denominator. We also saw that it is a bit difficult to carry the work in our head as the numbers grow larger. In Chapter 1, Section 3, we saw that canceling reduces the size of the numbers and simplifies the work. You Try It! Simplify: EXAMPLE 4. Simplify: 72 ( -x 64 1 -x Solution: In Example 3, we used our calculator to multiply 72 and 8 to get 576, then divided 576 by 9 to get 64. In this solution, we divide 9 into 72 to get 8, then multiply 8 by 8 to get 64. We get the same answer, but because the intermediate numbers are much smaller, the calculations are much easier to do mentally. 2.3. CLEARING FRACTIONS AND DECIMALS 101 72I-* 72 • - )x 9, (8-8)x 64cc Associative property of multiplication. Divide: 72/9 Multiply: 8 ■ I 64. Answer: 4Cte Example 4 shows all of the steps involved in arriving at the answer. Again, the goal in this section is to perform this calculation mentally, so we just "Divide 9 into 72 to get 8, then multiply 8 by 8 to get 644." □ 72,-* 64x Not only does this approach allow us to write down the answer without doing any work, the numerical calculations involve smaller numbers. You should practice this mental calculation until you can write down the answer without writing down any steps. You Try It! EXAMPLE 5. Simplify: 27 I -x Solution: Divide 9 into 27 to get 3, then multiply 3 by 5 to get 15. 27 (-a; | = I'm Simplify: l l X Answer: 27a; Note: The technique shown in Examples 4 o-nd 5 is the technique we '11 use in the remainder of this section. Dividing (canceling) first is far more efficient, the smaller numbers allowing us to perform the calculation mentally. Clearing Fractions from an Equation Now that we've done the required fraction work, we can now concentrate on clearing fractions from an equation. Once the fractions are removed from the equation, the resulting equivalent equation is far easier to solve than the orig- inal. □ Clearing fractions from an equation. To clear fractions from an equation, multiply both sides of the equation by the least common denominator. 102 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES You Try It! Solve for x: 3 _ 1 4 ~~ 2 EXAMPLE 6. Solve for 2 _ 1 3 ~ 2' Solution: The common denominator for 2/3 and 1/2 is 6. We begin by multiplying both sides of the equation by 6. x 2 _ 1 3 ~ 2 X+ l)= 6 {2 °' + ' 5 «i)= 6 U Original equation. Multiply both sides by 6. On the left, distribute the 6. To simplify 6(2/3), divide 6 by 3 to get 2, then multiply 2 by 2 to get 4. Thus, 6(2/3) = 4. Similarly, 6(1/2) = 3. Qx + 4 = 3 1 1) =4 > 6 u) Note that the fractions are now cleared from the equation. To isolate terms containing x on one side of the equation, subtract 4 from both sides of the equation. 6x + 4-4 = 3-4 6a; = -1 Subtract 4 from both sides. Simplify both sides. To "undo" multiplying by 6, divide both sides by 6. 6x _ -1 ~6~ ~ ir 1 X = ~6 Divide both sides by 6. Simplify both sides. Check: Let's use the TI-84 to check the solution. 1. Store —1/6 in the variable X using the following keystrokes. EDQHH ENTER The result is shown in the first line in Figure 2.9. 2. Enter the left-hand side of the original equation: x + 2/3. Use the fol- lowing keystrokes. 2.3. CLEARING FRACTIONS AND DECIMALS 103 X,T,0,n 0B0 ENTER The result is shown in the second line in Figure 2.9. 3. Press the MATH key, then select l:^Frac and press the ENTER key. The result is shown in the third line in Figure 2.9. -l/6-»X flns^Frac .5 1/2 Figure 2.9: Checking that —1/6 is a solution of x + 2/3 = 1/2. The result is identical to the right-hand side of the equation x + 2/3 = 1/2. Thus, the solution checks. Answer: x = 5/4 □ EXAMPLE 7. Solve for x: 4 — x 5 You Try It! Solve for Solution: The common denominator for 4/5 and —4/3 is 15. We begin by multiplying both sides of the equation by 15. l.-,||, 15 Original equation. Multiply both sides by 15. To simplify 15(4/5), divide 5 into 15 to get 3, then multiply 3 by 4 to get 12. Thus, 15(4/5) = 12. Similarly, 15(-4/3) = -20 12a; = -20 Multiply. To "undo" multiplying by 12, we divide both sides by 12. Divide both sides by 12. Reduce to lowest terms. 12a; -20 12 12 5 x = ~3 3 3 "7 X= 2 104 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Check: To check, substitute —5/3 for x in the original equation. 5 5 3 20 ~15 4 Original equation. Substitute —5/3 for x. Multiply numerators and denominators. Reduce. Answer: x -7/2 The fact that the last line is a true statement guarantees the —5/3 is a solution 4 4 of the equation —x = . 4 5 3 □ You Try It! Solve for 5x 2 _ 5 3x ~9~ ~ 3 ~ 9 ~ T EXAMPLE 8. Solve for x: —-- = --—. 3 4 2 4 Solution: The common denominator for 2a;/3, —3/4, 1/2, and — 3x/4 is 12. We begin by multiplying both sides of the equation by 12. 2x 3 1 3a; 12 12 3 4 2x 3 Y~ 4 12 ■ 3 2 12 12 4 1 3a; 2 ~ T 1 12 Original equation. Multiply both sides by 12. Distribute the 12 on each side. To simplify 12(2a;/3), divide 3 into 12 to get 4, then multiply 4 by 2a; to get 8a;. Thus, 12(2x/3) = 8x. Similarly, 12(3/4) = 9, 12(1/2) = 6, and 12(3x/4) = 9a;. 8x - 9 = 6 - 9x Multiply. Note that the fractions are now cleared from the equation. We now need to isolate terms containing x on one side of the equation. To remove the term —9a; from the right-hand side, add 9a; to both sides of the equation. 8a; - 9 + 9a; = 6 - 9a; + 9a; 17a; - 9 = 6 Add 9a; to both sides. Simplify both sides. To remove the term —9 from the left-hand side, add 9 to both sides of the equation. 17a;- 9 + 9 = 6 + 9 17a; = 15 Add 9 to both sides. Simplify both sides. 2.3. CLEARING FRACTIONS AND DECIMALS 105 Finally, to "undo" multiplying by 17, divide both sides of the equation by 17. 17a: _ 15 TT ~ 17 15 17 X Divide both sides by 17. Simplify both sides. Check: Let's use the TI-84 to check the solution. 1. Store the number 15/17 in the variable X using the following keystrokes. Q0BQQ X,T,0,n I ENTER The result is shown in Figure 2.10. .8823529412 Figure 2.10: Storing 15/17 in X. 2x 3 2. Enter the left-hand side of the original equation: . Use the fol- S H 3 4 lowing keystrokes. QBQBQ ENTER The result is shown in the first image in Figure 2.11). 1 3a; 3. Enter the right-hand side of the original equation: — . Use the following keystrokes. □ BEBH □ ENTER The result is shown in the second image in Figure 2.11. Because both sides simplify to —.1617647059 when 15/17 is substituted for x, this guarantees that 15/17 is a solution of the equation 2x/3 — 3/4 = 1/2 — 3a;/4. Answer: x = 22/37 □ 106 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 2*X/3-3/4 -.1617647059 l/2-3*X/4 -.1617647059 Figure 2.11: Checking 15/17 in 2a;/3 - 3/4 = 1/2 - 3x/4. Clearing Decimals from an Equation Multiplying by the appropriate power of ten will clear the decimals from an equation, making the resulting equivalent equation far easier to solve. Before we begin, first recall the following facts about multiplying by powers of ten. You Try It! Solve for x: 1.34- 4.5a; = 2.2 • 10(1.2345) = 12.345. Multiplying by 10 moves the decimal point one place to the right. • 100(1.2345) = 123.45. Multiplying by 100 moves the decimal point two places to the right. • 1000(1.2345) = 1234.5. Multiplying by 1000 moves the decimal point three places to the right. Note the pattern: The number of zeros in the power of ten determines the number of places to move the decimal point. So, for example, if we multiply by 1,000,000, which has six zeros, this will move the decimal point six places to the right. EXAMPLE 9. Solve for 2.3a; - 1.25 = 0.04a;. Solution: The first term of 2.3a; — 1.25 = 0.04a; has one decimal place, the second term has two decimal places, and the third and final term has two decimal places. At a minimum, we need to move each decimal point two places to the right in order to clear the decimals from the equation. Consequently, we multiply both sides of the equation by 100. 2.3a;- 1.25 100(2. 3a;- 1.25) 100(2. 3a;) - 100(1.25) 230a; - 125 0.04a; Original equation. 100(0.04x) Multiply both sides by 100. 100(0.04a;) Distribute the 100. 4a; Multiplying by 100 moves all decimal points two places to the right. 2.3. CLEARING FRACTIONS AND DECIMALS 107 Note that the decimals are now cleared from the equation. We must now isolate all terms containing x on one side of the equation. To remove the term Ax from the right-hand side, subtract Ax from both sides of the equation. 230a; — 125 — Ax = Ax — Ax Subtract Ax from both sides. 226a; - 125 = Simplify. To remove —125 from the left-hand side, add 125 to both sides of the equation. 226a; - 125 + 125 = + 125 Add 125 to both sides. 226a; = 125 Simplify both sides. Finally, to "undo" multiplying by 226, divide both sides by 226. 226a: 125 Divide both sides by 226. Simplify. Check: Let's check the answer with the TI-84. 1. Store 125/226 in the variable X using the following keystrokes. 226 226 x = 125 226 Q00H000 ENTER The result is shown in the first image in Figure 2.12. 2. Enter the left-hand side of the original equation: 2.3a: — 1.25. Use the following keystrokes. 0D0I BBB0000 ENTER The result is shown in the second image in Figure 2.12. 3. Enter the right-hand side of the original equation: 0.04a;. Use the follow- ing keystrokes. mnmEDBB The result is shown in the third image in Figure 2.12. 108 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 125/226-*X .5530973451 2.3*X-1.25 .0221238938 0.04*X ,0221238938 Figure 2.12: Checking 125/226 in 2.3a; - 1.25 = 0.04s. Answer: x = -43/225 Note that both sides yield the same decimal approximation 0.0221238938 when 125/226 is substituted for x. This guarantees that 125/226 is a solution of 2.3a;- 1.25 = 0.04a;. □ 2.3. CLEARING FRACTIONS AND DECIMALS 109 ti. ;». ;». Exercises ■*j ■*-: •*; In Exercises 1-6, simplify the expression. 9 1. !.(.( -.t 2. 27 ( -a; 3. 14 | -x 4. -12 ( -a; 5. 70 ( -x 6. -27 [ -x In Exercises 7-18, for each of the following equations, clear fractions by multiplying both sides by the least common denominator. Solve the resulting equation for x and reduce your answer to lowest terms. 7. 9 1 ~7 X ~ 3 5 ~ 3 8. 1 3 ~2 X ~ 4 5 ~ ~9 9. 7 5 -x+ - = 3 9 2 4 — a; 3 3 10. 2 9 3^" 4 ~ 5 4 "S^" 3 11. 9 8 4 X "7 = 3 2 12. 7 2 ~3 X ~ 9 4 ~3 13. 4 3 14. 2 5 "j^" 7 15. 3 6 4 5 16. 2 1 X ~ 9 ~~ 4 17. 1 4 "j}^" 3 " 3 8 " 5 , o 6 3 9 1 18. — a; = — x 7 5 7 2 In Exercises 19-32, clear decimals from the given equation by multiplying by the appropriate power of ten, then solve the resulting equation for x. Your final answer should be a fraction reduced to lowest terms. 19. 2.39a; + 0.71 = -1.98a; + 2.29 20. 0.12a; + 0.52 = -1.47a; - 2.12 21. 0.4a;- 1.55 = 2.14 22. 0.8a; -2.18 = 1.49 23. 2.6a; -2.54 = -2.14a; 24. -1.4a; -2.98 = 0.55a; 25. 0.7a; = -2.3a; - 2.8 26. 3.4a; = 1.8s + 2.5 27. -4.8a; -2.7= -1.9 28. -2.4a; + 2.5 = 2.3 29. 1.7a; + 2.1 = -1.6a; + 2.5 30. -1.2a; + 0.4 = -2.7a; - 1.9 31. 2.5a; + 1.9 = 0.9a; 32. 4.4a; + 0.8 = 2.8x 1. 72a; 3. 21a; 5. 90a: 7 14 9 q 17 15 11. 74 63 13. 32 15. 9 20 17. 16 ~25 110 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES m- j*- j*- Answers •*$ •** •** 19. 158 437 21. 369 23. 127 237 25. 14 ~15 27. 1 ~6 29. 4 3.3 31. 19 ~16 2.4. FORMULAE 111 2.4 Formulae The formulae of science usually contain variable letters other than the variable x. Indeed, formulae in science typically use several letters. For example, Isaac Newton's Universal Law of Gravitation says that the magnitude of the force of attraction between two celestial bodies is given by the formula GMm where inn usually denotes the mass of the smaller body, M the mass of the larger body, and r is the distance between the two bodies. The letter G represents the universal gravitational constant, having value 6.67428 x 10~ n N(m/kg) . Variable case. Note the use of upper and lower case letter M's in Newton's Law of Gravitation. When working with scientific formulae, you must maintain the case of the given letters. You are not allowed to substitute lower for upper case, or upper for lower case in your work. In Section 2.2, we described the goal that must be met when we are asked to "solve an equation for ic." Solve for x. When asked to solve an equation for x, the goal is to manipulate the equation into the final form x= "Stuff", where "Stuff" is a valid mathematical expression that may contain other vari- ables, mathematical symbols, etc., but it must not contain any occurrence of the variable x. 'Formulae" is the plural for 'formula." Thus, to solve an equation for x, we need to isolate the terms containing x on one side of the equation, and all remaining terms on the other side of the equation. EXAMPLE 1. Solve for a;: x + a = b. Solution: To undo the effects of adding a, subtract a from both sides of the equation. x + a = b x + a — a = b — a x = b — a Original equation. Subtract a from both sides. Simplify. You Try It! Solve for x: x — c = d Answer: x = c + d. □ You Try It! Solve for x: x — c = d Answer: c = x — d. 112 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES In Example 1, note that the answer x = b — a has the required form, x = "Stuff" , where "Stuff" is a valid mathematical expression that contains other variables, mathematical symbols, etc., but it does not contain any occurrence of the variable x. Now, what if we were asked to solve the same equation for a, instead of xl EXAMPLE 2. Solve for a: x + a = b. Solution: We are instructed to solve the equation x + a = b for a. This means that our final answer must have the form a = "Stuff" , where "Stuff" is a valid mathematical expression that contains other variables, mathematical symbols, etc., but it does not contain any occurrence of the variable a. This means that we must isolate all terms containing the variable a on one side of the equation, and all remaining terms on the other side of the equation. Now, to undo the effect of adding x, subtract x from both sides of the equation. x + a = b x + a — x = b — x a = b — x Note that we have a = "Stuff" variable we are solving for. Original equation. Subtract x from both sides. Simplify. where "Stuff" contains no occurrence of a, the a You Try It! Solve for m: E = mc EXAMPLE 3. The formula F = kx, known as "Hooke's Law" , predicts the force F required to stretch a spring x units. Solve the equation for k. Solution: We are instructed to solve the equation F = kx for k. This means that our final answer must have the form k = "Stuff" , where "Stuff" is a valid mathematical expression that may contain other variables, mathematical symbols, etc., but it may not contain any occurrence of the variable k. This means that we must isolate all terms containing the variable k on one side of the equation, and all remaining terms on the other side of the equation. However, note that all terms containing the variable k are already isolated on one side of the equation. Terms not containing the variable k are isolated on the other side of the equation. Now, to "undo" the effect of multiplying by x, divide both sides of the equation by x. Original equation. Divide both sides by x. Simplify. F = kx F kx X X F 2.4. FORMULAE 113 Saying that F/x = k is equivalent to saying that k = F/x. We can leave our answer in the form shown in the last step, but some instructors insist that we write the answer as follows: k F x F/x = k is equivalent to k = F/x. Note that we have k = "Stuff" , where "Stuff" contains no occurrence of k, the variable we are solving for. Answer: m E □ EXAMPLE 4. The formula V = RI is called "Ohm's Law." It helps calculate the voltage drop V across a resistor R in an electric circuit with current /. Solve the equation for R. Solution: We are instructed to solve the equation V = RI for R. This means that our final answer must have the form R = "Stuff" , where "Stuff" is a valid mathematical expression that may contain other variables, mathematical symbols, etc., but it may not contain any occurrence of the variable R. This means that we must isolate all terms containing the variable R on one side of the equation, and all remaining terms on the other side of the equation. However, note that all terms containing the variable R are already isolated on one side of the equation. Terms not containing the variable R are isolated on the other side of the equation. Now, to "undo" the effect of multiplying by I, divide both sides of the equation by /. V = RI V _RI 7 ~ T v 7 R Original equation. Divide both sides by I. Simplify. This can also be written in the following form: R V 7 v/i. Note that we have R = "Stuff the variable we are solving for. V/I = R is equivalent to R ' , where "Stuff" contains no occurrence of 7?, You Try It! Solve for t: si Answer: t d s □ Clearing Fractions If fractions occur in a formula, clear the fractions from the formula by multi- plying both sides of the formula by the common denominator. 114 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES You Try It! Solve for g: s = -gV EXAMPLE 5. The formula K = —mv 2 is used to calculate the kinetic energy K of a particle of mass m moving with velocity v. Solve the equation for m. Solution: We're asked to solve the equation K = (l/2)mv 2 for m. First, clear the fractions by multiplying both sides by the common denominator. K = —mv 2{K) = 2 ( -mv 2 2K = mv 2 Original equation. Multiply both sides by 2. Simplify. Cancel 2's. Note that all terms containing m, the variable we are solving for, are already isolated on one side of the equation. We need only divide both sides by v 2 to complete the solution. 2s Answer: q = -r- t 2 2K mv 2 v 2 2K ?.' 2 v m Divide both sides by v . Simplify. Cancel v for v . Note that the final answer has the form m = "Stuff" , where "Stuff" contains no occurrence of the variable m. □ You Try It! Solve for q^. F kqiq 2 EXAMPLE 6. As mentioned earlier, Newton's Universal Law of Gravitation is described by the formula „ GMm Solve this equation for m. Solution: We're asked to solve the equation F = GMm/r 2 for m. First, clear the fractions by multiplying both sides by the common denominator. GMr w-*(2p) r 2 F = GMm Original equation. Multiply both sides by r . Simplify. Cancel r for r . 2.4. FORMULAE 115 Note that all terms containing m, the variable we are solving for, are already isolated on one side of the equation. We need only divide both sides by GM to complete the solution. r 2 F GMr GM r 2 F GM GM m Divide both sides by GM. Simplify. Cancel GM for GM. Note that the final answer has the form m no occurrence of the variable m. "Stuff" , where "Stuff" contains Answer: </2 = ~, Fr 2 kqi □ Geometric Formulae Let's look at a few commonly used formulae from geometry. EXAMPLE 7. Let W and L represent the width and length of a rectangle, respectively, and let P represent its perimeter. You Try It! The perimeter of a rectangle is 160 meters and its width is 30 meters. Finds its length. W IT" The perimeter (distance around) of the rectangle is found by summing its four sides, then combining like terms. p=L+W+L+W P = 2W + 2L Summing the four sides. Combine like terms. Problem: Solve P = 2W + 2L for L. Then, given that the perimeter is 300 feet and the width is 50 feet, use your result to calculate the length. Solution: We're first asked to solve P = 2W + 2L for L. First, isolate all 116 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES terms containing the variable L on one side of the equation. Original equation. Subtract 2W from both sides. Simplify. Divide both sides by 2. Simplify. z Note that the final result has L = "Stuff', where "Stuff" contains no occurrence of the variable L. The second part of this example requests that we find the length of the rectangle, given that the perimeter is P = 300 feet and the width is W = 50 feet. To calculate the length, substitute P = 300 and W = 50 in L = [P — 2W)/2. p = 2W + 2L p -2W = 2W + 2L- -2W p -2W = 2L p - -2W 2L p- 2 -2W 2 - L P-2W L = 2 L = 300- 2(50) 2 L = 300- 100 2 L = 200 ~2~ L = 100 Perimeter formula solved for L. Substitute 300 for P, 50 for W. Multiply: 2(50) = 100. Subtract: 300 - 100 = 200. Divide: 200/2 = 100. Answer: L = 50 meters Hence, the length of the rectangle is 100 feet. □ You Try It! The area of a triangle is EXAMPLE 8. Let 6 and h represent the length of the base and the height 140 cm 2 and the length of its of a triangle, respectively, and let A represent the area of the triangle, base is 70 cm. Find the height of the triangle. h The area of the triangle is computed using the formula: A = -bh 2 2.4. FORMULAE 117 That is, the area of a triangle is "one-half the base times the height." Problem: Solve the formula A = ^bh for h. Secondly, given that the area is A = 90 in" (90 square inches) and the length of the base is 15 in (15 inches), find the height of the triangle. Solution: We're first asked to solve A = (l/2)bh for h. Because the equation has fractions, the first step is to clear the fractions by multiplying both sides by the least common denominator. A = — bh Area of a triangle formula. 2 2(A) = 2 ( - bh ) Multiply both sides by 2. 2A = bh Simplify. Now, we already have all terms containing the variable h on one side of the equation, so we can solve for h by dividing both sides of the equation by b. 2A bh — = — Divide both sides by o. b b —— = h Simplify. Note that the final result has h = "Stuff', where "Stuff" contains no occurrence of the variable h. The second part of this example requests that we find the height of the triangle, given that the area is A = 90 in and the length of the base is b = 15 in. To calculate the height of the triangle, substitute A = 90 and b = 15 in h = 2A/b. 94 Area formula solved for h. Substitute 90 for A, 15 for b. Multiply: 2(90) = 180. Divide: 180/15 = 12. Hence, the height of the triangle is 12 inches. Answer: 4 cm h = 2A ~b h = 2(90) 15 h = 180 h = 12 □ 118 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES ;».;-». t%.. Exercises •*;■*; -*i In Exercises 1-30, solve the given formulas for the indicated variable. 1. F = kx for x 2. A = irr 2 for ir mc 2 for ?7, 4. v = Vq + at for vq 5. A = -nr\ri for r 2 6. y = mx + b for b 7. F = ma for a 8. V = Iwh for I 9. C = 2irr for r 10. F = kx for x 11. y = mx + b for x V 12. I = — for i? ii 13. F = gwB for i> 14. x + <Z = e for a; 1 15. V 'h for /i 16. P = IRT for I V 17. I = - for 7? 18. F = -C + 32 for C 5 19. F — =- for g 20. Ax + By = C for y 21. P = 2W + 2L for VF 22. A = -h(bt + b 2 ) for h \ h ^ a + b + c 23. A 24. A: 25. y- 26. A: 27. F 28. A: 3 29. d = uf for u 30. x + d = e for d for ft, 3 y = m(x - xo) for m —bh for b 2 GA f for M a + b + c or h 31. Let W and L represent the width and length of a rectangle, respectively, and let A represent its area. W L W The area of the rectangle is given by the formula A = LW. Solve this formula for L. Then, given that the area of the rectangle is A = 1073 square meters and its width is W = 29 meters, determine its length. 32. Let b\ and b 2 represent the parallel bases of a trapezoid and let h represent its height. 2.4. FORMULAE 119 The area of the trapezoid is given by the formula i(* b 2 )h. Solve this formula for b\. Then, given that the area is A = 2457 square centimeters, the second base is b 2 = 68 centimeters, and the height is h = 54 centimeters, find the length b\ of the first base. 33. A parallelogram is a quadrilateral (four- sided figure) whose opposite sides are par- allel. The area of the parallelogram is computed using the formula: bh Solve this formula for b. Next, given that the area is A = 2418 square feet and the height is h = 31 feet, find the length of the base of the parallelogram. 34. Let b\ and b 2 represent the parallel bases of a trapezoid and let h represent its height. 35 The area of the trapezoid is given by the formula A=^(b 1 +b 2 )h. Solve this formula for h. Then, given that the area is A = 3164 square yards, the bases are b\ = 38 yards and 62 = 75 yards, find the height h of the trapezoid. Let b and h represent the length of the base and the height of a triangle, respec- tively, and let A represent the area of the triangle. The area of the triangle is computed using the formula: A=-bh 2 Solve this formula for b. Next, given that the area is A = 1332 square inches and the height is h = 36 inches, find the length of the base of the triangle. 36. The circumference of a circle, somewhat like the term perimeter, is the distance around the circle. The diameter of a cir- cle is a line segment drawn through the center of the circle. Since the time of the ancient Greeks, it has been known that the ratio of the cir- cumference to the diameter is a constant, denoted by the symbol w. C 7 = 7r 120 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Solve the formula for d. Then, given C = 188.4 yards and tt = 3.14, find the length of the diameter d. 37. Let W and L represent the width and length of a rectangle, respectively, and let P represent its perimeter. L W W L The perimeter (distance around) of the rectangle is found by summing its four sides, then combining like terms. P = 2W + 2L Solve P = 2W + 2L for W. Then, given that the perimeter is P = 256 meters and the length is L = 73 meters, use your re- sult to calculate the width. j*- j*- j*- Answers •*$-*$■*$ F E 3. m = — ; cr 19. q 21. W r 2 F ~kQ P-2L 5. r 2 7. a F m 9. r = — 2tt 11. x 13. v 15. h 17. R y-b m F qB W ■KT 2 V 23. h 25. m 27. M : 2A 6i + 6 2 _ y-yo X — Xo r 2 F G 111 d 29. v = - t 31. 37 square meters 33. 78 feet 35. 74 inches 37. 55 meters 2.5. APPLICATIONS 121 2.5 Applications The solution of a word problem must incorporate each of the following steps. Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. You must let your readers know what each variable in your problem represents. This can be accomplished in a number of ways: • Statements such as "Let P represent the perimeter of the rectangle." • Labeling unknown values with variables in a table. • Labeling unknown quantities in a sketch or diagram. 2. Set up an Equation. Every solution to a word problem must include a carefully crafted equation that accurately describes the constraints in the problem statement. 3. Solve the Equation. You must always solve the equation set up in the previous step. 4. Answer the Question. This step is easily overlooked. For example, the problem might ask for Jane's age, but your equation's solution gives the age of Jane's sister Liz. Make sure you answer the original question asked in the problem. Your solution should be written in a sentence with appropriate units. 5. Look Back. It is important to note that this step does not imply that you should simply check your solution in your equation. After all, it's possible that your equation incorrectly models the problem's situation, so you could have a valid solution to an incorrect equation. The important question is: "Does your answer make sense based on the words in the original problem statement." Let's give these requirements a test drive. You Try It! EXAMPLE 1. Three more than five times a certain number is —62. Find 27 more than 5 times a the number. certain number is —148. Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. Let x represent the unknown number. 2. Set up an Equation. "Three more than five times a certain number is —62" becomes: 122 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Three more than five times a certain number 5.T -62 -62 3 + 5x = -62 + 5x - 3 = -62- -3 5x = -65 5x -65 ~~5 ~ 5 x = -13 Answer: -35 3. Solve the Equation. To solve for x, first subtract 3 from both sides of the equation. Original equation. Subtract 3 from both sides. Simplify. Divide both sides by 5. Simplify. 4. Answer the Question. The unknown number is —13. 5. Look Back. Compute "three more than five times —13." 3 + 5(-13) = 3 + (-65) = -62 Hence, three more than five times —13 is —62, as required. Our solution is correct. □ You Try It! The sum of three consecutive odd integers is —225. What are the integers? EXAMPLE 2. The sum of three consecutive integers is smallest of these three integers. -66. Find the Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1 . Set up a Variable Dictionary. Let k represent the smallest of three con- secutive integers. 2. Set up an Equation. An example of three consecutive integers is 34, 35, and 36. These are not the integers we seek, but they serve to help in the understanding of the problem. Note how each consecutive integer is one larger than the preceding integer. If k is the smallest of three consecutive integers, then the next two consecutive integers are k + 1 and k + 2. The "sum of three consecutive integers is —66" becomes: 2.5. APPLICATIONS 123 First second third consecutive plus consecutive plus consecutive integer integer integer + (fc + 1) (fc + 2) -66 -66 3. Solve the Equation. To solve for k, first simplify the left-hand side of the equation by combining like terms. fc + (fc + l) + (fc + 2) = -66 Original equation. 3fc + 3 = -66 Combine like terms. 3fc+3-3= -66-3 Subtract 3 from both sides 3fc = -69 Simplify. 3fc y ~ -69 3 Divide both sides by 3. k = -23 Simplify. 4. Answer the Question. The smallest of three consecutive integers is —23. 5. Look Back. If k = — 23 is the smallest of three consecutive integers, then the next two consecutive integers are —22 and —21. Let's check the sum of these three consecutive integers. -23 +(-22) + (-21) -66 Hence, the sum of the three consecutive integers is —66, as required. Our solution is correct. Answer: —77, —75, —73 □ You Try It! EXAMPLE 3. A carpenter cuts a board measuring 60 inches in three pieces. The second piece is twice as long as the first piece, and the third piece is three times as long as the first piece. Find the length of each piece cut by the carpenter. Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. Let L represent the length of the first piece. Then the second piece, which is twice as long as the first piece, has length 2L. The third piece, which is three times as long as the first piece, has length 3L. Let's construct a little table to help summarize the information provided in this problem. Han cuts a board measuring 230 inches in three pieces. The second piece is twice as long as the first piece, and the third piece is 30 inches longer than the second piece. Find the length of each piece cut by Han. 124 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Piece Length (in) First piece Second piece Third piece L 2L 3L Total length 60 2. Set up an Equation. As you can see in the table above, the second column shows that the sum of the three pieces is 60 inches. In symbols: L + 2L + 3L = 60 3. Solve the Equation. To solve for L, first simplify the left-hand side of the equation by combining like terms. L 2L + 3L = 60 6L = 60 6L _ 60 it ~ y L= 10 Original equation. Combine like terms. Divide both sides by 6. Simplify. 4. Answer the Question. The first piece has length L = 10 inches. The second piece has length 2L = 20 inches. The third piece has length 3L = 30 inches. In tabular form, this is even more apparent. Piece Length (in) Length (in) First piece Second piece Third piece L 2L 3L 10 20 30 Total length 60 60 Answer: 40, 80, 110 in 5. Look Back. Not only is the second length twice the first and the third length three times the first, check the sum of their lengths: 10 + 20 + 30 = 60 That's a total of 60 inches. We have the correct solution. □ You Try It! The three sides of a triangle are consecutive integers. If the perimeter (sum of the three sides) of the triangle is 453 centimeters, find the length of each side of the triangle. EXAMPLE 4. The three sides of a triangle are consecutive even integers. If the perimeter (sum of the three sides) of the triangle is 156 centimeters, find the length of each side of the triangle. 2.5. APPLICATIONS 125 Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. An example of three consecutive even integers is 18, 20, and 22. These are not the integers we seek, but they do give us some sense of the meaning of three consecutive even integers. Note that each consecutive even integer is two larger than the preceding integer. Thus, if k is the length of the first side of the triangle, then the next two sides are fc + 2 and fc+4. In this example, our variable dictionary will take the form of a well-labeled figure. k + 2 2. Set up an Equation. The perimeter of the triangle is the sum of the three sides. If the perimeter is 156 centimeters, then: fc + (fc + 2) + (fc + 4) = 156 3. Solve the Equation. To solve for k, first simplify the left-hand side of the equation by combining like terms. k + ( fc + 2) + (fc + 4) = ■■ 156 Original equation. 3fc + 6 = : 156 Combine like terms. 3fc+6-6= 156-6 Subtract 6 from both sides 3fc = : 150 Simplify. 3fc y ~ 150 Divide both sides by 3. k = 50 Simplify. 4. Answer the Question. Thus, the first side has length 50 centimeters. Because the next two consecutive even integers are k + 2 = 52 and fc + 4 = 54, the three sides of the triangle measure 50, 52, and 54 centimeters, respectively. 5. Look Back. An image helps our understanding. The three sides are consecutive even integers. 126 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 54 cm 50 cm 52 cm Note that the perimeter (sum of the three sides) is: 50 cm + 52 cm + 54 cm = 156 cm (2.1) Answer: 150, 151, 152 cm Thus, the perimeter is 156 centimeters, as it should be. Our solution is correct. □ You Try It! The second angle of a triangle is three times bigger than the first angle. The third angle of the triangle is a 40 degrees larger than the second angle. How many degrees are in each angle? EXAMPLE 5. A well-known fact from geometry is the fact that the sum of the angles of a triangle is 180°. Suppose we have a triangle whose second angle is 10 degrees larger than twice its first angle and whose third angle is 50 degrees larger than its first angle. Find the measure of each angle of the triangle. Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. The Greek alphabet starts out with the letters a, j3, 7, 5, e, . . . , in much the same way that the English alphabet start out with the letters a, b, c, d, e, . . . . Mathematicians love to use Greek letters, especially in the study of trigonometry. The greek letter 9 (pronounced "theta" ) is particularly favored in representing an angle of a triangle. So, we'll let 6 represent the degree measure of the first angle of the triangle. The second angle is 10 degrees larger than twice the first angle, so the second angle is 26 + 10. The third angle is 50 degrees larger than the first angle, so the third angle is 9 + 50. Again, we'll set up a well-labeled figure for our variable dictionary. 2.5. APPLICATIONS 127 2. Set up an Equation. The sum of the angles is 180°, so: + (20 + 10) + (0 + 50) = 180 3. Solve the Equation. To solve for 0, first simplify the left-hand side of the equation by combining like terms. + (20 + 10) + (0 + 50) = 180 Original equation. 40 + 60 = 180 Combine like terms. 40 + 60 - 60 = 180 - 60 Subtract 60 from both sides. 40 = 120 Simplify. 40 _ 120 T ~~ ~T = 30 Simplify. Divide both sides by 4. 4. Answer the Question. Thus, the first angle is = 30 degrees, the second angle is 20 + 10 = 70 degrees, and the third angle is + 50 = 80 degrees. 5. Look Back. An image helps our understanding. Note that the second angle is 10 degrees larger than twice the first angle. Note that the third angle is 50 degrees larger than the first angle. Note that the sum of the angles is: 30° + 70° + 80° = 180° (2.2) Thus, the sum of the three angles is 180 degrees, as it should be. We have the correct solution. Answer: 20°, 60°, 100° □ EXAMPLE 6. Martha inherits $21,000 and decides to invest the money in three separate accounts. The amount she invests in the second account is twice the amount she invests in the first account. The amount she invests in You Try It! Jim inherits $15,000. He invests part in a fund that pays 5% per year and the rest in a fund that pays 4% per year. At the end of one year, the combined interest from both investments was $4,250. How much did he invest in each fund? 128 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES the third account is $1,000 more than the amount she invests in the second account. How much did she invest in each account? Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. We'll use a table in this example to help set up our variable dictionary. Let x be the amount invested in the first account. The amount invested in the second account is twice that invested in the first account, so 2x is the amount invested in the second account. The third account investment is $1,000 more than the amount invested in the second account, so that is 2x + 1000. Account # Amount Invested Account #1 Account #2 Account #3 X 2x 2x+ 1000 Total Invested 21000 2. Set up an Equation. The second column of the table reveals the required equation. The three investments must sum to $21,000. x + 2x+{2x + 1000) = 21000 3. Solve the Equation. To solve for x, first simplify the left-hand side of the equation by combining like terms. x + 2x + (2x + 1000) = 21000 Original equation. 5x + 1000 = 21000 Combine like terms. 5x + 1000 - 1000 = 21000 - 1000 Subtract 1000 from both sides. 5a; = 20000 Simplify. 5x _ 20000 ~5 ~ 5 x = 4000 Simplify. Divide both sides by 5. 4. Answer the Question. Substitute x = 4000 in each entry of the second column of the table above to produce the results in the table below. Account # Amount Invested Amount Invested Account # 1 Account #2 Account^ 1,000 S,000 ),000 Total Invested 21000 $21,000 2.5. APPLICATIONS 129 Look Back. As we can see in our answer table, the amount $8,000 invested in the second account is twice the amount invested in the first account. The amount $9,000 invested in the third account is %1,000 more than the amount invested in the second account. Moreover, the total investment is: $4, 000 + $8, 000 + $9, 000 = $21, 000 (2.3) Thus, the total investment is $21,000, as it should be. We have the correct solution. Answer: $5,000 at 5% and $10,000 at 4%. □ You Try It! EXAMPLE 7. Jeff is hiking the 2,650-mile Pacific Crest Trail from Mexico to Canada. Shortly before he crosses over from Oregon into Washington he is four times as far from the beginning of the trail as he is from the end. How much further does he have to hike? Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. Let d represent the distance left for Jeff to hike. Because Jeff is four times further from the beginning of the trail than the end, the distance Jeff already completed is Ad. Let's construct a little table to help summarize the information provided in this problem. Margaret is cycling along a lane that measures 100 miles. If Magaret is four times as far from the start of the ride as she is from the finish, how many more miles does she have to go before she finishes her ride? Section of Trail Distance (mi) Distance to finish Distance from start d Ad Total distance 2650 2. Set up an Equation. As you can see in the table above, the second column shows that the sum of the two distances is 2650 miles. In symbols: d + 4d= 2650 3. Solve the Equation. To solve for d, first simplify the left-hand side of the equation by combining like terms. d + Ad = 2650 5d = 2650 5d _ 2650 ~5~ ~~ 5 d = 530 Original equation. Combine like terms. Divide both sides by 5. Simplify. 130 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 4. Answer the Question. Jeff still has 530 miles to hike. 5. Look Back. Because the amount left to hike is d = 530 miles, Jeff's distance from the start of the trail is 4rf = 4(530), or 2,120 miles. If we arrange these results in tabular form, it is evident that not only is the distance from the start of the trail four times that of the distance left to the finish, but also the sum of their lengths is equal to the total length of the trail. Section of Trail Distance (mi) Distance (mi) Distance to finish Distance from start d Ad 530 2120 Total distance 2650 2650 Answer: 20 miles Thus, we have the correct solution. □ You Try It! 20% of Mary's class were ill and stayed home from school. If only 36 students are present, what is the actual size of Mary's class? EXAMPLE 8. Today 15% of Sister Damaris' seventh grade class were ill and stayed home from school. If only 34 students are present, what is the actual size of Sister Damaris' class? Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. Let S represent the actual size of Sister Damaris' class. 2. Set up an Equation. If 15% of Sister Damaris' class was absent, then 85% of her class was present. There are 34 student present, so the phrase "85% of Sister Damaris' class is 34" translates into the equation, 0.85,5 = 34, where we've changed 85% to a decimal by moving the decimal point two places to the left. 3. Solve the Equation. To solve for S, first clear the decimals by multiplying both sides of the equation by 100. 0.855 = 34 85,5 = 3400 85S* _ 3400 "85" ~ 85 S* = 40 Original equation. Multiply both sides by 100. Divide both sides by 85. Simplify. 2.5. APPLICATIONS 131 4. Answer the Question. Sister Damaris' class size is 40. 5. Look Back. We're told that 15% of Sister Damaris' class is absent. If we calculate 15% of 40, we get: 0.15(40) = 6 Thus, there were 6 students absent, so 40 — 6, or 34 students were present. Thus, we have the correct solution. Answer: 45 □ 132 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES f»- n- fa- Exercises ■*1 ■*'» .*5 1. Two angles are said to be complementary if their sum is ninety degrees. Suppose you have two complementary angles such that the second angle is 6 degrees larger than 2 times the measure of the first an- gle. Find the angles. 2. Two angles are said to be complementary if their sum is ninety degrees. Suppose you have two complementary angles such that the second angle is 10 degrees larger than 3 times the measure of the first an- gle. Find the angles. 3. Two angles are said to be supplementary if their sum is 180 degrees. Suppose you have two supplementary angles such that the second angle is 10 degrees larger than 4 times the measure of the first angle. Find the angles. 4. Two angles are said to be supplementary if their sum is 180 degrees. Suppose you have two supplementary angles such that the second angle is 12 degrees larger than 5 times the measure of the first angle. Find the angles. 5. The three sides of a triangle are consecu- tive integers. If the perimeter (sum of the three sides) of the triangle is 483 meters, find the length of each side of the triangle. 6. The three sides of a triangle are consecu- tive integers. If the perimeter (sum of the three sides) of the triangle is 486 yards, find the length of each side of the trian- gle. 7. Four less than eight times a certain num- ber is —660. Find the number. 8. Nine less than five times a certain number is 141. Find the number. 9. Alan is hiking a trail that is 70 miles long. After several days, he is four times as far from the beginning of the trail as he is from the end. How much further does he have to hike? 10. Joe is hiking a trail that is 30 miles long. After several days, he is two times as far from the beginning of the trail as he is from the end. How much further does he have to hike? 11. Martha takes roll in her sixth grade grade class and finds that 2 students are missing. If her actual class size is 36 students, what percentage of her class is absent? Round your answer to the nearest percent. 12. Alice takes roll in her first grade grade class and finds that 7 students are missing. If her actual class size is 37 students, what percentage of her class is absent? Round your answer to the nearest percent. 13. Lily cuts a piece of yarn into three pieces. The second pieces is 3 times as long as the first piece, and the third piece is 6 cen- timeters longer than the first piece. If the total length of the yarn is 211 centime- ters, find the lengths of each of the three pieces. 14. Jane cuts a piece of twine into three pieces. The second pieces is 7 times as long as the first piece, and the third piece is 5 feet longer than the first piece. If the total length of the twine is 320 feet, find the lengths of each of the three pieces. 15. The three sides of a triangle are consecu- tive even integers. If the perimeter (sum of the three sides) of the triangle is 450 yards, find the length of each side of the triangle. 2.5. APPLICATIONS 133 16. The three sides of a triangle are consecu- tive even integers. If the perimeter (sum of the three sides) of the triangle is 318 feet, find the length of each side of the triangle. 17. The perimeter of a triangle is 414 yards. The second side of the triangle is 7 times as long as the first side and the third side of the triangle is 9 yards longer than the first side. Find the lengths of each of the three sides of the triangle. 18. The perimeter of a triangle is 54 inches. The second side of the triangle is 2 times as long as the first side and the third side of the triangle is 6 inches longer than the first side. Find the lengths of each of the three sides of the triangle. 19. The sum of three consecutive odd integers is —543. Find the smallest of the three consecutive odd integers. 20. The sum of three consecutive odd integers is —225. Find the smallest of the three consecutive odd integers. 21. The sum of the angles of a triangle is 180°. In triangle AABC, the degree measure of angle B is 4 times the degree measure of angle A. The degree measure of angle C is 30 degrees larger than the degree measure of angle A. Find the degree measures of each angle of triangle AABC. 22. The sum of the angles of a triangle is 180°. In triangle AABC, the degree measure of angle B is 4 times the degree measure of angle A. The degree measure of angle C is 60 degrees larger than the degree measure of angle A. Find the degree measures of each angle of triangle AABC. 23. The sum of three consecutive integers is — 384. Find the largest of the three con- secutive integers. 24. The sum of three consecutive integers is — 501. Find the largest of the three con- secutive integers. 25. Seven more than two times a certain num- ber is 181. Find the number. 26. Nine more than two times a certain num- ber is 137. Find the number. 27. The three sides of a triangle are consecu- tive odd integers. If the perimeter (sum of the three sides) of the triangle is 537 feet, find the length of each side of the triangle. 28. The three sides of a triangle are consecu- tive odd integers. If the perimeter (sum of the three sides) of the triangle is 471 centimeters, find the length of each side of the triangle. 29. A store advertises that it is offering a 14% discount on all articles purchased at the store. If Yao pays $670.80 for an article, what was the marked price for the article? 30. A store advertises that it is offering a 12% discount on all articles purchased at the store. If Roberto pays $560.56 for an ar- ticle, what was the marked price for the article? 31. The sum of three consecutive even integers is —486. Find the smallest of the three consecutive even integers. 32. The sum of three consecutive even integers is —354. Find the smallest of the three consecutive even integers. 33. Burt inherits $45,500. He decides to in- vest part of the inheritance in a mutual fund and the remaining part in a certifi- cate of deposit. If the amount invested in the certificate of deposit is $3,500 more than 6 times the amount invested in the mutual fund, find the amount invested in each account. 34. Phoenix inherits $12,000. He decides to invest part of the inheritance in a mutual fund and the remaining part in a certifi- cate of deposit. If the amount invested in the certificate of deposit is $3,000 more than 8 times the amount invested in the mutual fund, find the amount invested in each account. 134 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES $*■ j*- &*■ Answers •*$ *** •** 1. 28, 62 degrees 19. -183 3. 34, 146 degrees 21. 25°, 100°, 55° 5. 160, 161, 162 meters 03 —127 7. -82 9. 14 miles 11. 6% 13. 41, 123, and 47 centimeters 15. 148, 150, 152 yards 17. 45, 315, and 54 yards 25. 87 27. 177, 179, 181 feet 29. $780.00 31. -164 33. $6,000, $39,500 2.6. INEQUALITIES 135 2.6 Inequalities In Chapter 1, we introduced the natural numbers N = {1, 2,3,.. .}, the whole numbers W = {0, 1, 2, 3, . . .}, and the integers Z = {. . . , -3, -2, -1, 0, 1, 2, 3, . . .}. Later in the chapter, we introduced the rational numbers, numbers of the form p/q, where p and q are integers. We noted that both terminating and repeating decimals are rational numbers. Each of these numbers has a unique position on the number line (see Figure 2.13). -3.125 0.3 1.5 .'52 -4-3-2-10 1 2 3 4 Figure 2.13: Positioning numbers on the number line. The natural numbers, whole numbers, and integers are also rational num- bers, because each can be expressed in the form p/q, where p and q are integers. For example, 0=0/12, 4 = 4/1, and —3 = —12/4. Indeed, the rational numbers contain all of the numbers we've studied up to this point in the course. However, not all numbers are rational numbers. For example, consider the decimal number —3.10110111011110 . . ., which neither terminates nor repeats. The number v2 = 1.414213562373095 . . . also equals a decimal number that never terminates and never repeats. A similar statement can be made about the number 7r = 3.141592653589793 . . .. Each of these irrational (not rational) numbers also has a unique position on the number line (see Figure 2.14). -3.010110111.. —i w 1- V2 7T -4-3-2-10 1 2 3 4 Figure 2.14: Positioning numbers on the number line. The Real Numbers. If we combine all of the rational and irrational numbers into one collection, then we have a set of numbers that is called the set of real numbers. The set of real numbers is denoted by the symbol K. Two other irrational numbers you may encounter in your mathematical studies are e (Euler's constant), which is approximately equal to ew 2.71828182845904..., and (pronounced "phi" ) , called the golden ratio, which equals cp = (1 + v / 5)/2. The number e arises in applications involving compound interest, probability, and other areas of mathematics. The number 4> is used in financial markets and is also arguably the ratio of beauty in art and architecture. Every point on the number line is associated with a unique real number. Con- versely, every real number is associated with a unique position on the number line. In lieu of this correspondence, the number line is usually called the real line. 136 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Ordering the Real Numbers The real numbers are ordered on the real line in a manner identical to how we ordered the integers on the number line in Section 1 of Chapter 1 . Order on the Real Line. Suppose that a and b are real numbers positioned on the real line as shown below. • Because a lies to the "left of" b, we say that a is "less than" b, or in mathematical symbols, a < b. The inequality symbol < is read "less than." • Alternately, b lies to the "right of" a, so we can also say that b is "greater than" a, or in mathematical symbols, b > a. The inequality symbol > is read "greater than." Here are two more inequality symbols that we will use in this section. Less than or equal to. If we want to say that a lies to the "left of" 6, or shares the same position as 6, then we say that a is "less than or equal to" b and write a < b. The inequality symbol < is pronounced "less than or equal to." Greater than or equal to. If we want to say that b lise to the "right of" a, or shares the same position as a, then we say that b is "greater than ore equal to o and write b > a. The inequality symbol > is pronounced "greater than or equal to." Set-Builder Notation Mathematicians use a construct called set-builder notation to describe sets or collections of numbers. The general form of set-builder notation looks as follows: {x : some statement about x} For example, suppose that we want to describe the set of "all real numbers that are less than 2." We could use the following notation: {x : x < 2} 2.6. INEQUALITIES 137 This is read aloud as follows: U A equals the set of all x such that x is less than 2." Some prefer to use a vertical bar instead of a colon. A = {x\x < 2} In this text we use the colon in set-builder notation, but feel free to use the vertical bar instead. They mean the same thing. One might still object that the notation {x : x < 2} is a bit vague. One objection could be "What type of numbers x are you referring to? Do you want the integers that are less than two or do you want the real numbers that are less than two?" As you can see, this is a valid objection. One way of addressing this objection is to write: A = {ieK:i<2) or A = {x e N : x < 2} The first is read "A is the set of all x in K that are less than two," while the second is read "A is the set of all x in N that are less than two." Set-builder Assumption. In this text, unless there is a specific refernce to the set of numbers desired, we will assume that the notation {x : x < 2} is asking for the set of all real numbers less than 2. In Figure 2.15, we've shaded the set of real numbers {x : x < 2}. Because ^— I 1 1 1 1 1 1 O 1 1 1 — > -5-4-3-2-10 1 2 3 4 5 Figure 2.15: Shading the numbers less than 2. "less than" is the same as saying "left of," we've shaded (in red) all points on the real line that lie to the left of the number two. Note that there is an "empty circle" at the number two. The point representing the number two is not shaded because we were only asked to shade the numbers that are strictly less than two. While the shading in Figure 2.15 is perfectly valid, much of the information provided in Figure 2.15 is unnecessary (and perhaps distracting). We only need to label the endpoint and shade the real numbers to the left of two, as we've done in constructing Figure 2.16. < 3 > 2 Figure 2.16: You only need to label the endpoint. For contrast, suppose instead that we're asked to shade the set of real numbers {x : x < 2}. This means that we must shade all the real numbers 138 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Figure 2.17: You only need to label the endpoint. You Try It! Shade {x : x < 4} on the real line. Answer: that are "less than or equal to 2" or "left of and including 2." The resulting set is shaded in Figure 2.17. Note the difference between Figures 2.16 and 2.17. In Figures 2.16 we're shading the set {x : x < 2}, so the number 2 is left unshaded (an empty dot). In Figures 2.17, we're shading the set {x : x < 2}, so the number 2 is shaded (a filled- in dot). EXAMPLE 1. Shade the set {x : x > -3} on the real line. Solution: The notation {x : x > —3} is pronounced "the set of all real numbers x such that x is greater than or equal to —3." Thus, we need to shade the number —3 and all real numbers to the right of —3. <- -> □ You Try It! Use set-builder notation to describe the following set of real numbers: O- -10 ■¥ Answer: {x : x > —10} EXAMPLE 2. Use set-builder notation to describe the set of real numbers that are shaded on the number line below. «- O Solution: The number —1 is not shaded. Only the numbers to the left of —1 are shaded. This is the set of all real numbers x such that x is "less than" —1. Thus, we describe this set with the following set-builder notation: {•-• x < ■1} □ Interval Notation In Examples 1 and 2, we used set-builder notation to describe the set of real numbers greater than or equal to —3 and a second set of real numbers less than — 1. There is another mathematical symbolism, called interval notation, that can be used to describe these sets of real numbers. Consider the first set of numbers from Example 1, {x : X > —3}. 2.6. INEQUALITIES 139 Sweeping our eyes "from left to right", we use [— 3,oo) to describe this set of real numbers. Some comments are in order: 1. The bracket at the left end means that —3 is included in the set. 2. As you move toward the right end of the real line, the numbers grow without bound. Hence, the oo symbol (positive infinity) is used to indi- cate that we are including all real numbers to the right of —3. However, oo is not really a number, so we use a parentheses to indicate we are "not including" this fictional point. The set of numbers from Example 2 is {a: : x < —I}. < o > -1 Sweeping our eyes "from left to right" , this set of real numbers is described with (— oo, —1). Again, comments are in order: 1. The number —1 is not included in this set. To indicate that it is not included, we use a parenthesis. 2. As you move toward the left end of the real line, the numbers decrease without bound. Hence, the — oo symbol (negative infinity) is used to indicate that we are including all real numbers to the left of —1. Again, — oo is not an actual number, so we use a parenthesis to indicate that we are not including this "fictional" point. Sweep your eyes from "left to right" . If you would like to insure that you correctly use interval notation, place the numbers in your interval notation in the same order as they are encountered as you sweep your eyes from "left to right" on the real line. A nice summary of set-builder and interval notation is presented in Table 2.1 at the end of the section. Equivalent Inequalities Like equations, two inequalities are equivalent if they have the same solution sets. 140 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Adding or Subtracting the Same Quantity from Both Sides of an Inequality. Let a and b be real numbers with If c is any real number, then and a < b, a + c < b + c c < b — c. That is, adding or subtracting the same amount from both sides of an inequality produces an equivalent inequality (does not change the solution). You Try It! Use interval notation to describe the solution of: x-7 < -8 Answer: -oo,—l] EXAMPLE 3. Solve for a;: x - 2 < 7. Sketch the solution on the real line, then use set-builder and interval notation to describe your solution. Solution: To "undo" subtracting 2, we add 2 to both sides of the inequality. x — 2 < 7 Original inequality. s-2 + 2<7 + 2 Add 2 to both sides. x < 9 Simplify both sides. To shade the real numbers less than or equal to 9, we shade the number 9 and all real numbers to the left of 9. < • > 9 Using set-builder notation, the solution is {x : x < 9}. Using interval notation, the solution is (— oo,9]. □ If we multiply or divide both sides of an inequality by a positive number, we have an equivalent inequality. Multiplying or Dividing by a Positive Number. Let a and b be real numbers with a < b. If c is a rea] positive number, then ac < be and a b - < -. c c 2.6. INEQUALITIES 141 EXAMPLE 4. Solve for x: 3x < —9 Sketch the solution on the real line, then use set-builder and interval notation to describe your solution. Solution: To "undo" multiplying by 3, divide both sides of the inequality by 3. Because we are dividing both sides by a positive number, we do not reverse the inequality sign. You Try It! Use interval notation to describe the solution of: 2x > -8 3x < -9 Original inequality. 3ir — < 3 ~ -9 ~3~ Divide both sides by 3 x < -3 Simplify both sides. Shade the real numbers less than or equal to —3. Using set-builder notation, the solution is {x : x < —3}. Using interval nota- tion, the solution is (— oo, —3]. Answer: (—4, oo) Reversing the Inequality Sign Up to this point, it seems that the technique for solving inequalities is pretty much identical to the technique used to solve equations. However, in this section we are going to encounter one exception. Suppose we start with the valid inequality — 2 < 5, then we multiply both sides by 2, 3, and 4. □ -2 < 5 -2 < 5 -2 < 5 2(-2) < 2(5) 3(-2) <3(5) 4(-2) < 4(5) -4< 10 -6 < 15 -8 < 20 Note that in each case, the resulting inequality is still valid. Caution! We're about to make an error! Start again with — 2 < 5, but this time multiply both sides by —2, —3, and —4. -2 < 5 -2 < 5 -2 < 5 -2(-2) < -2(5) -3(-2)<-3(5) -4(-2) < -4(5) 4<-10 6<-15 8< -20 142 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Note that in each of the resulting inequalities, the inequality symbol is pointing the wrong way! Some readers might prefer a more formal reason as to why we reverse the inequality when we multiply both sides by a negative number. Suppose that a < b. Then, subtracting b from both sides gives the result a — b < 0. This means that a — b is a negative number. Now, if c is a negative number, then the product (a — b)c is positive. Then: (a-b)c> ac — be > ac — be + be > + be ac > be Thus, if you start with a < b and c < 0, then ac > be. You Try It! Use interval notation to describe the solution of: -3a; > -6 When you multiply both sides of an inequality by a negative number, you must reverse the inequality sign. Starting with — 2 < 5, multiply both sides by —2, —3, and —4, but reverse the inequality symbol. -2 < 5 -2(-2) > -2(5) 4> -10 -2 < 5 -3(-2) > -3(5) 6 > -15 -2 < 5 -4(-2) > -4(5) 8 > -20 Note that each of the resulting inequalities is now a valid inquality. Multiplying or Dividing by a Negative Number. Let a and b be real numbers i with a < b. If c is a real negative number, ac > be then and a b - > -. c c That is, when multiplying or dividing both sides of an inequality by a negative number you must reverse the inequality sign. EXAMPLE 5. Solve for x: — 2x < 4. Sketch the solution on the real line, then use set-builder and interval notation to describe your solution. Solution: To "undo" multiplying by —2, divide both sides by —2. Because we are dividing both sides by a negative number, we reverse the inequality sign. -2x <4 -2x 4 ^2 > ~2 x > Original inequality. Divide both sides by —2. Reverse the inequality sign. Simplify both sides. Shade the real numbers greater than —2. -O Answer: (— oo,2] Using set-builder notation, the solution is {x : x > — 2}. Using interval nota- tion, the solution is (— 2,oo). □ 2.6. INEQUALITIES 143 Multiple Steps Sometimes you need to perform a sequence of steps to arrive at the solution. EXAMPLE 6. Solve for x: 2x + 5 > -7. Sketch the solution on the real line, then use set-builder and interval notation to describe your solution. Solution: To "undo" adding 5, subtract 5 from both sides of the inequality. 2x + 5 > — 7 Original inequality. 2a; + 5 — 5>— 7— 5 Subtract 5 from both sides. 2a; > — 12 Simplify both sides. To "undo" multiplying by 2, divide both sides by 2. Because we are dividing both sides by a positive number, we do not reverse the inequality sign. 2.r -12 > x > Divide both sides by 2. Simplify both sides. Shade the real numbers greater than —6. -O- -6 You Try It! Use interval notation to describe the solution of: 3a; - 2 < 4 Using set-builder notation, the solution is {x : x > —6}. Using interval nota- tion, the solution is (—6, oo). Answer: (— oo,2] □ EXAMPLE 7. Solve for x: 3 - 5a: < 2a: + 17. Sketch the solution on the real line, then use set-builder and interval notation to describe your solution. Solution: We need to isolate terms containing x on one side of the inequality. Start by subtracting 2a; from both sides of the inequality. 3- 5a; < 2a; + 17 5a; - 2a: < 2a; + 17 - 3 -7a; < 17 Original inequality. 2a; Subtract 2a; from both sides. Simplify both sides. We continue to isolate terms containing x on one side of the inequality. Subtract 3 from both sides. 3- 7a;- 3 < 17 -7a; < 14 Subtract 3 from both sides. Simplify both sides. You Try It! Use interval notation to describe the solution of: 4 - x > 2x + 1 144 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES To "undo" multiplying by —7, divide both sides by — 7. Because we are dividing both sides by a negative number, we reverse the inequality sign. -Ix 14 — > — -7 " -7 x > -2 Shade the real numbers greater than or equal to —2 Divide both sides by —7. Simplify both sides. Answer: (—00, 1) You Try It! Use interval notation to describe the solution of: 2x T > Using set-builder notation, the solution is {x : x > — 2}. Using interval nota- tion, the solution is [— 2,oo). □ We clear fractions from an inequality in the usual manner, by multiplying both sides by the least common denominator. EXAMPLE 8. Solve for x_ 1 12 > 3' Solution: First, clear the fractions from the inequality by multiplying both sides by the least common denominator, which in this case is 12. x 1 12 > 3 12 3 x 4 ~ 12 > '1" 3 "3' 4_ 1.12- > T 3 9- a; >4 Original inequality. 12 Multiply both sides by 12. 12 Distribute the 12. Cancel and multiply. To "undo" adding 9, subtract 9 from both sides. 9-2-9 >4-9 -x > -5 Subtract 9 from both sides. Simplify both sides. We could divide both sides by —1, but multiplying both sides by —1 will also do the job. Because we are multiplying both sides by a negative number, we reverse the inequality sign. (-l)(-x) < (-5)(-l) x < 5 Multiply both sides by —1. Reverse the inequality sign. Simplify both sides. 2.6. INEQUALITIES 145 Shade the real numbers less than 5. i -O- 5 Using set-builder notation, the solution is {x : x < 5}. Using interval notation, the solution is (— oo, 5). Answer: [— 9/8, oo) We clear decimals from an inequality in the usual manner, by multiplying both sides by the appropriate power of ten. □ EXAMPLE 9. Solve for x: 3.25 - 1.2a; > 4.6. Solution: First, clear the decimals from the inequality by multiplying both sides by 100, which moves each decimal point two places to the right. 3.25- 1.2cc > 4.6 325 - 120a; > 460 325 - 120a; - 325 > 460 - 325 -120a; > 135 -120a; 135 < -120 x < -120 27 24 Original inequality. Multiply both sides by 100. Subtract 325 from both sides. Simplify both sides. Divide both sides by —120. Reverse the inequality sign. Reduce to lowest terms. Shade the real numbers less than —27/24. «- O -27/24 Using set-builder notation, the solution is {x : x < —27/24}. Using interval notation, the solution is (-co, —27/24). You Try It! Use interval notation to describe the solution of: 2.3a; -5.62 > -1.4 Answer: [211/115, oo) □ 146 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES Summary Table of Set-Builder and Interval Notation A summary table of the set-builder and interval notation is presented in Table 2.1. Shading on the real I ine Set-builder Interval {x : x > —5} -5 (-5,oo) _^ {x : x > —5} -5 ^ [-5,oo) — > {x : x < —5} -5 (-oo,-5) / A {x : x < —5} -5 (-oo,-5] Table 2.1: Examples of set-builder and interval notation. 2.6. INEQUALITIES 147 ti. ;». ;». Exercises ■*j ■*-: •*; 1. Draw a number line, then plot the num- bers 4, 3, —4, 7/8, and —8/3 on your num- ber line. Label each point with its value. Finally, list the numbers in order, from smallest to largest. 2. Draw a number line, then plot the num- bers 5, 3, —4, 5/7, and —4/3 on your num- ber line. Label each point with its value. Finally, list the numbers in order, from smallest to largest. Draw a number line, then plot the num- bers — 5, 5, 4, 2/3, and 8/3 on your num- ber line. Label each point with its value. Finally, list the numbers in order, from smallest to largest. Draw a number line, then plot the num- bers — 3, —2, 4, 1/3, and 5/2 on your num- ber line. Label each point with its value. Finally, list the numbers in order, from smallest to largest. In Exercises 5-20, shade each of the following sets on a number line. 5. {x : x > -7} 6. {x : x > -1} 7. {x : x < 2} 8. {x : x < -6} 9. (-oo,2) 10. (-oo,-9) 11. (6,00) 12. (5,oo) 13. {x : x > 7} 14. {x : x > -8} 15. [0,oo) 16. [7,oo) 17. {x : x < -2} 18. {x : x < 7} 19. (-oo,3] 20. (-oo,-l] In Exercises 21-28, use set-builder notation to describe the shaded region on the given number line. 21. <- 22. «- 23. <- 24. <- 9 -•- ■O -o- 9 25. 26. 27. 28. -O -o 148 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES In Exercises 29-36, use interval notation to describe the shaded region on the given number line. 29. < O > 33. «- 4 30. < O > 34. «- -7 31. < O > 35. <- 32. < O > 36. 2 In Exercises 37-44, solve each of the given inequalities. Sketch the solution on a number line, then use set-builder and interval notation to describe your solution. 37. a; + 10< 19 41. -2x < -2 38. x + 17 > 7 42. -18a; > -20 39. Ax < 8 43. x- 18 > -10 40. 16a; > -2 44. x - 8 < -18 In Exercises 45-62, solve each of the given inequalities. Sketch the solution on a number line, then use set-builder and interval notation to describe your solution. 45. -5a; -6 > 4 - 9a; 54. -11a; - 7 > -15 - 5x 46. 2a;- 7 > -3- 4a; 55. 2a; - 9 > 5 - 8a; 47. 16a;- 6 < 18 56. -3a; - 6 > -2 - 9a; 48. 8a;- 14 < -12 57. -10a; - 4 < 18 49. -14a;- 6 > -10- 4a; 58. -6a;- 14 <1 50. -13a;- 4 > -2- 5a; 59. -12a; + 4 < -56 51. 5a; + 18 < 38 60. -18a; + 6 < -12 52. 9a; + 16 < 79 61. 15a; + 5 < 6a; + 2 53. -16a;- 5 > -11 -6a; 62. 12a; + 8 < 3a; + 5 2.6. INEQUALITIES 149 In Exercises 63-76, solve each of the given inequalities. Sketch the solution on a number line, then use set-builder and interval notation describe your solution. 63. 3 9 -x > - 2 8 64. 6 3 7 X > 7 7 4 65. 3 9 X+ 2<5 66. 1 1 x +l < ~7 4 5 67. 4 1 4 x < —x - 7 6~3 1 " 2 68. 5 3 7 x < —x - 3 4-4 3 " 5 69. 3 9 x > — 8 " 7 70. 7 1 a? > - 2-5 71. 6 4 5^-7 72. 4 2 -x < - 3-9 73. 6 7 5 2 — x < X 5 3-99 74. 3 13 2 — x < X 7 2~2 7 75. 9 9 17 —x H — > —x H — 7 2 7 2 76. 5 9 15 —x H — > —x H — 7 2 3 2 In Exercises 77-84, solve each of the given inequalities. Sketch the solution on a number line, then use set-builder and interval notation containing fractions in reduced form to describe your solution. 77. -3.7a;- 1.98 < 3.2 78. -3.6a; -3.32 < 0.8 79. -3.4a; + 3.5 > 0.9 - 2.2.x 80. -2.6a; + 3.1 > -2.9- 1.7x 81. -1.3a; + 2.9 > -2.6 - 3.3x 82. 2.5x + 2.1 > 1.4 -3.8a; 83. 2.2x + 1.9 < -2.3 84. 1.6a: +1.2 < 1.6 j*- j*- f*- Answers •*$ ■*$ ■** 3. 5. 3 4 -5-4-3-2-10 1 2 3 4 5 3 4 5 — •-! • • > -5-4-3-2-10 1 2 3 4 5 9. <- 11. <- 13. <- 15. <- O 2 -O 6 -O 7 7. <- O 2 17. <- 150 CHAPTER 2. SOLVING LINEAR EQUATIONS AND INEQUALITIES 19. <- 21. {x : x < 9} 23. {x : x < -8} 25. {x : x > -2} 27. {x : x > —3} 29. (4,oo) 31. (-oo,-2) 33. (— oo, 5] 35. [l,oo) 37. (-oo,9) 39. (-oo,2) 41. [l,oo) 43. (8,00) 45. [5/2,oo) 47. (-oo,3/2] 49. [2/5,oo) 51. (-oo,4) 53. 3/5, 00) 55. 7/5,oo) 57. -ll/5,oo) 59. 5, 00) 61. -oo,-l/3) 63. 3/4,oo) 65. -oo,3/10) 67. 5/7,oo) 69. -51/56, 00) 71. 10/21,oo) 73. -65/22, 00) 75. -7/8,oo) 77. -7/5,oo) 79. -oo,13/6] 81. -ll/4,oo) 83. -oo,-21/ll Chapter 3 Introduction to Graphing In thinking about solving algebra and geometry problems in the 1600's, Rene Descartes came up with the idea of using algebraic procedures to help solve geometric problems and to also demonstrate algebraic operations by using ge- ometry. One of the easy ways to do this was to use the 2-dimensional geometric plane. He decided to use the letters near the end of the alphabet to name the axes: "x-axis" for the horizontal axis and "y-axis" for the vertical axis. Then ordered pairs (x, y) would be used to graph specific points on the plane and/or algebraic expressions and equations. Thus, the graphs on the following pages are done on a graph named a "Cartesian Coordinate" graph in Rene Descartes' honor. The equation of a particular line can be written in several algebraic forms which you will learn to interpret. The geometric aspect of a line will be noted in its position and slope. We will also be focusing on geometric aspects such as parallel and perpendicular lines. Ultimately, these tasks will be used to help solve application problems involving speed and distance. In modern times, we use the graphing calculator to speed up the process of providing tables of data, graphing lines quickly, and finding solutions of equations. 151 152 CHAPTER 3. INTRODUCTION TO GRAPHING 3.1 Graphing Equations by Hand We begin with the definition of an ordered pair. Ordered Pair. The construct (x,y), where x and y are any real numbers, is called an ordered pair of real numbers. (4,3), (—3,4), (—2,-3), and (3,-1) are examples of ordered pairs. Order Matters. Pay particular attention to the phrase "ordered pairs." Or- der matters. Consequently, the ordered pair (x, y) is not the same as the ordered pair (y,x), because the numbers are presented in a different order. Rene Descartes (1596-1650) was a French philosopher and mathematician who is well known for the famous phrase "cogito ergo sum" (I think, therefore I am), which appears in his Discours de la methode pour bien conduire sa raison, et chercher la verite dans les sciences (Discourse on the Method of Rightly Conducting the Reason, and Seeking Truth in the Sciences). In that same treatise, Descartes introduces his coordinate system, a method for representing points in the plane via pairs of real numbers. Indeed, the Cartesian plane of modern day is so named in honor of Rene Descartes, who some call the "Father of Modern Mathematics" The Cartesian Coordinate System Pictured in Figure 3.1 is a Cartesian Coordinate System. On a grid, we've created two real lines, one horizontal labeled x (we'll refer to this one as the x-axis), and the other vertical labeled y (we'll refer to this one as the y-axis). y -5 -4 -3 -2 -1 ->■ x 12 3 4 5 -1 -2 -3 -4- -5- Figure 3.1: The Cartesian Coordinate System. Two Important Points: Here are two important points to be made about the horizontal and vertical axes in Figure 3.1. 3. 1 . GRAPHING EQUATIONS BY HAND 153 1. As you move from left to right along the horizontal axis (the x-axis in Figure 3.1), the numbers grow larger. The positive direction is to the right, the negative direction is to the left. 2. As you move from bottom to top along the vertical axis (the y-axis in Figure 3.1), the numbers grow larger. The positive direction is upward, the negative direction is downward. Additional Comments: Two additional comments are in order: 3. The point where the horizontal and vertical axes intersect in Figure 3.2 is called the origin of the coordinate system. The origin has coordinates (0,0). 4. The horizontal and vertical axes divide the plane into four quadrants, numbered I, II, III, and IV (roman numerals for one, two, three, and four), as shown in Figure 3.2. Note that the quadrants are numbered in a counter-clockwise order. II / Origin: (0,0) 5 - 4 - 3 - 2 - Tl "T 1J .1 IV Figure 3.2: Numbering the quadrants and indicating the coordinates of the origin. 154 CHAPTER 3. INTRODUCTION TO GRAPHING Plotting Ordered Pairs Before we can plot any points or draw any graphs, we first need to set up a Cartesian Coordinate System on a sheet of graph paper? How do we do this? What is required? We don't always label the horizontal axis as the x-axis and the vertical axis as the y-axis. For example, if we want to plot the velocity of an object as a function of time, then we would be plotting points (t, v). In that case, we would label the horizontal axis as the t-axis and the vertical axis as the u-axis. Draw and label each axis. If we are going to plot points (x,y), then, on a sheet of graph paper, perform each of the following initial tasks. 1. Use a ruler to draw the horizontal and vertical axes. 2. Label the horizontal axis as the x-axis and the vertical axis as the y-axis. The result of this first step is shown in Figure 3.3. Indicate the scale on each axis. 3. Label at least one vertical gridline with its numerical value. 4. Label at least one horizontal gridline with its numerical value. The scales on the horizontal and vertical axes may differ. However, on each axis, the scale must remain consistent. That is, as you count to the right from the origin on the x-axis, if each gridline represents one unit, then as you count to the left from the origin on the x-axis, each gridline must also represent one unit. Similar comments are in order for the y-axis, where the scale must also be consistent, whether you are counting up or down. An example is shown in Figure 3.4. Note that the scale indicated on the x-axis indicates that each gridline counts as 1-unit as we count from left-to-right. The scale on the y-axis indicates that each gridlines counts as 2-units as we count from bottom-to-top. .'/ ■*■ x -10 -10- * x Figure 3.3: Draw and label each axis. Figure 3.4: Indicate the scale on each axis. 3. 1 . GRAPHING EQUATIONS BY HAND 155 Now that we know how to set up a Cartesian Coordinate System on a sheet of graph paper, here are two examples of how we plot points on our coordinate system. To plot the ordered pair (4, 3), start at the origin and move 4 units to the right along the horizontal axis, then 3 units upward in the direction of the vertical axis. (4 3) To plot the ordered pair (—2,-3), start at the origin and move 2 units to the left along the horizontal axis, then 3 units downward in the direc- tion of the vertical axis. y (-2,-3) Continuing in this manner, each ordered pair (x, y) of real numbers is asso- ciated with a unique point in the Cartesian plane. Vice-versa, each point in the Cartesian point is associated with a unique ordered pair of real numbers. Because of this association, we begin to use the words "point" and "ordered pair" as equivalent expressions, sometimes referring to the "point" (x, y) and other times to the "ordered pair" (x,y). EXAMPLE 1. Identify the coordinates of the point P in Figure 3.5. Solution: In Figure 3.6, start at the origin, move 3 units to the left and 4 units up to reach the point P. This indicates that the coordinates of the point P are (-3,4). You Try It! Identify the coordinates of the point P in the graph below. y V 5 c. / ) 5 ■ 156 CHAPTER 3. INTRODUCTION TO GRAPHING Answer: (3,-2) P{=3^) ->■ x *■ x Figure 3.5: Identify the coordinates of the point P. Figure 3.6: Start at the origin, move 3 units left and 4 units up. □ The variables do not have to always be x and y. For example, the equation v = 2 + 3.2£ is an equation in two variables, v and t. Equations in Two Variables The equation y = x + 1 is an equation in two variables, in this case x and y. Consider the point (x, y) = (2,3). If we substitute 2 for x and 3 for y in the equation y = x + 1, we get the following result: x 2- 3 Original equation. Substitute: 2 for x, 3 for y. Simplify both sides. Because the last line is a true statement, we say that (2, 3) is a solution of the equation y = x + 1. Alternately, we say that (2,3) satisfies the equation y = x + 1. On the other hand, consider the point (x, y) = (—3, 1). If we substitute —3 for x and 1 for y in the equation y = x + 1, we get the following result. y l 1 : x + 1 -3 + 1 -2 Original equation. Substitute: —3 for x, 1 for y. Simplify both sides. Because the last line is a false statement, the point (—3, 1) is not a solution of the equation y = x+1; that is, the point (—3, 1) does not satisfy the equation y = x + 1. 3. 1 . GRAPHING EQUATIONS BY HAND 157 Solutions of an equation in two variables. Given an equation in the variables x and y and a point (x, y) = (a, b), if upon subsituting a for x and b for y a true statement results, then the point (x, y) = (a, b) is said to be a solution of the given equation. Alternately, we say that the point (x, y) = (a, b) satisfies the given equation. You Try It! EXAMPLE 2. Which of the ordered pairs (0,-3) and (1,1) satisfy the Which of the ordered pairs equation y = 3x — 2? Solution: Substituting the ordered pairs (0, —3) and (1, 1) into the equation y = 3x — 2 lead to the following results: (—1,3) and (2, 1) satisfy the equation y = 2x + 5? Consider (x,y) = (0, —3). Substitute for x and —3 for y: y = 3 a: — 2 -3 = 3(0) -2 -3 = -2 The resulting statement is false. Consider (x, y) = (1, 1). Substitute 1 for x and 1 for y: y = 3a; — 2 1 = 3(1) -2 1 = 1 The resulting statement is true. 3a; — 2, but Thus, the ordered pair (0, —3) does not satisfy the equation y the ordered pair (1, 1) does satisfy the equation y = 3a; — 2. Answer: (—1,3) Graphing Equations in Two Variables Let's first define what is meant by the graph of an equation in two variables. The graph of an equation. The graph of an equation is the set of all points that satisfy the given equation. □ EXAMPLE 3. Sketch the graph of the equation y 1. Solution: The definition requires that we plot all points in the Cartesian Coordinate System that satisfy the equation y = x + 1. Let's first create a table of points that satisfy the equation. Start by creating three columns with headers x, y, and (a;, y), then select some values for x and put them in the first column. You Try It! Sketch the graph of the equation y = —x + 2. 158 CHAPTER 3. INTRODUCTION TO GRAPHING X y -■ = x + l (x,y) -3 -2 (-3,-2) -2 -1 1 2 3 Take the first value of x, namely x = —3, and substitute it into the equation y = x + 1. y = x + 1 y = -3 + i 2/ = -2 Thus, when x = —3, we have y = —2. Enter this value into the table. Continue substituting each tabular value of x into the equation y = x + 1 and use each result to complete the corresponding entries in the table. y = -3 + 1 = -2 2/=-2 + l = -l y = -1 + 1 = y = 0+1 = 1 y=l+l=2 y = 2 + 1 = 3 y = 3+1 =4 The last column of the table now contains seven points that satisfy the equation y = x+l. Plot these points on a Cartesian Coordinate System (see Figure 3.7). X y = x + 1 (&>?/) -3 -2 (-3,-2) -2 -1 (-2,-1) -1 (-1,0) 1 (0,1) 1 2 (1,2) 2 3 (2,3) 3 4 (3,4) y v Figure 3.7: Seven points that sat- isfy the equation y = x + 1. Figure 3.8: Adding additional points to the graph of y = X + 1. In Figure 3.7, we have plotted seven points that satisfy the given equation y = x + 1. However, the definition requires that we plot all points that satisfy the equation. It appears that a pattern is developing in Figure 3.7, but let's calculate and plot a few more points in order to be sure. Add the a;-values 3. 1 . GRAPHING EQUATIONS BY HAND 159 — 2.5, —1.5, —0.5, 05, 1.5, and 2.5 to the x-column of the table, then use the equation y = x + 1 to evaluate y at each one of these x-values. y = -2.5 + 1 = -1.5 y = -1.5 + 1 = -0.5 y = -0.5 + 1 =0.5 y = 0.5 + 1 = 1.5 y = 1.5 + 1 = 2.5 y = 2.5 + 1 = 3.5 X y = x + 1 (x, y) -2.5 -1.5 (-2.5,-1.5) -1.5 -0.5 (-1.5,-0.5) -0.5 0.5 (-0.5,0.5) 0.5 1.5 (0.5,1.5) 1.5 2.5 (1.5,2.5) 2.5 3.5 (2.5,3.5) Add these additional points to the graph in Figure 3.7 to produce the image shown in Figure 3.8. There are an infinite number of points that satisfy the equation y = x + 1. In Figure 3.8, we've plotted only 13 points that satisfy the equation. However, the collection of points plotted in Figure 3.8 suggest that if we were to plot the remainder of the points that satisfy the equation y = x + 1, we would get the graph of the line shown in Figure 3.9. Figure 3.9: The graph of y = x + 1 is a line. □ Guidelines and Requirements Example 3 suggests that we should use the following guidelines when sketching the graph of an equation. 160 CHAPTER 3. INTRODUCTION TO GRAPHING Guidelines for drawing the graph of an equation. When asked to draw the graph of an equation, perform each of the following steps: 1. Set up and calculate a table of points that satisfy the given equation. 2. Set up a Cartesian Coordinate System on graph paper and plot the points in your table on the system. Label each axis (usually x and y) and indicate the scale on each axis. 3. If the number of points plotted are enough to envision what the shape of the final curve will be, then draw the remaining points that satisfy the equation as imagined. Use a ruler if you believe the graph is a line. If the graph appears to be a curve, freehand the graph without the use of a ruler. 4. If the number of plotted points do not provide enough evidence to en- vision the final shape of the graph, add more points to your table, plot them, and try again to envision the final shape of the graph. If you still cannot predict the eventual shape of the graph, keep adding points to your table and plotting them until you are convinced of the final shape of the graph. Here are some additional requirements that must be followed when sketch- ing the graph of an equation. Graph paper, lines, curves, and rulers. The following are requirements for this class: 5. All graphs are to be drawn on graph paper. 6. All lines are to be drawn with a ruler. This includes the horizontal and vertical axes. 7. If the graph of an equation is a curve instead of a line, then the graph should be drawn freehand, without the aid of a ruler. Using the TABLE Feature of the Graphing Calculator As the equations become more complicated, it can become quite tedious to create tables of points that satisfy the equation. Fortunately, the graphing calculator has a TABLE feature that enables you to easily construct tables of points that satisfy the given equation. 3. 1 . GRAPHING EQUATIONS BY HAND 161 EXAMPLE 4. Use the graphing calculator to help create a table of points that satisfy the equation y = x 2 — 7 . Plot the points in your table. If you don't feel that there is enough evidence to envision what the final shape of the graph will be, use the calculator to add more points to your table and plot them. Continue this process until your are convinced of the final shape of the graph. Solution: The first step is to load the equation y = x 2 — 7 into the Y= menu of the graphing calculator. The topmost row of buttons on your calculator (see Figure 3.10) have the following appearance: You Try It! TRACE GRAPH WINDOW Pressing the Y= button opens the Y= menu shown in Figure 3.11. Use the up-and-down arrow keys (see Figure 3.13) to move the cursor after Yl= in the Y= menu, then use the following keystrokes to enter the equation y = x 2 — 7. The result is shown in Figure 3.12. mom ENTER Ploti Plots Plot3 \Vl = ■^E = ■^3 = \Yh = ■^Ye = ^Vfi = ^v?= Ploti Plots Plots ^iBX A 2-7 ■^£=1 ■^3 = \Yh = ■^Ye = ^Vfi = sV? = Figure 3.11: Opening the Y= menu. Figure 3.12: Entering the equation y = x 2 - 7. Figure 3.10: The graphing calculator. The next step is to "set up" the table. First, note that the calculator has symbolism printed on its case above each of its buttons. Above the WINDOW button you'll note the phrase TBLSET. Note that it is in the same color as the 2ND button. Thus, to open the setup window for the table, enter the following keystrokes. WINDOW Set TblStart (see Figure 3.14) equal to the first x- value you want to see in your table. In this case, enter —4 after TblStart. Set ATbl to the increment you want for your x- values. In this case, set ATbl equal to 1. Finally, set Figure 3.13: Arrow keys move the cursor. 162 CHAPTER 3. INTRODUCTION TO GRAPHING both the independent and dependent variables to "automatic." In each case, use the arrow keys to highlight the word AUTO and press ENTER. The result is shown in Figure 3.14. Next, note the word TABLE above the GRAPH button is in the same color as the 2ND key. To open the TABLE, enter the following keystrokes. GRAPH The result is shown in Figure 3.15. Note that you can use the up-and-down arrow keys to scroll through the table. TABLE SETUP TblStart=-4 *Tbl=I Indpnt: EH135 flsk Depend: rslffiK flsk Figure 3.14: Opening the TBLSET menu. X Vi -i 1 9 £ -fi -? -fi "3 Press + for ^Tbll Figure 3.15: Table of points satis- fying y = x 2 - 7. Next, enter the results from your calculator's table into a table on a sheet of graph paper, then plot the points in the table. The results are shown in Figure 3.16. V X y = x 2 - 7 (x,y) -4 9 (-4,9) -3 2 (-3,2) -2 -3 (-2,-3) -1 -6 (-1,-6) -7 (0,-7) 1 -6 (1,-6) 2 -3 (2,-3) 3 2 (3,2) 4 9 (4,9) 11 ) . 1( 1 1( ) 1 Figure 3.16: Plotting nine points that satisfy the equation y = x 2 — 7. In Figure 3.16, the eventual shape of the graph of y 7 may be evident already, but let's add a few more points to our table and plot them. Open the 3. 1 . GRAPHING EQUATIONS BY HAND 163 table "setup" window again by pressing 2ND WINDOW. Set TblStart to -4 again, then set the increment ATbl to 0.5. The result is shown in Figure 3.17. TABLE SETUP TblStart=-4 *Tbl=.5 Indpnt: [ilBE flsk Depend: Figure 3.17: Set ATbl equal to 0.5. X V1 -S.E -3 -£.£ -E -i.E -i 9 £.££ 2 -.?s "3 "H.FE -fi ku=h Figure 3.18: More points satisfying y = x 2 - 7. Add these new points to the table on your graph paper and plot them (see Figure 3.19). X y = x 2 - 7 (x,y) -3.5 5.25 (-3.5,5.25) -2.5 -0.75 (-2.5,-0.75) -1.5 -4.75 (-1.5,-4.75) -0.5 -6.75 (-0.5,-6.75) 0.5 -6.75 (0.5,-6.75) 1.5 -4.75 (1.5,-4.75) 2.5 -0.75 (2.5,-0.75) 3.5 5.25 (3.5,5.25) 1( M • • If • • 10 • • if p 1f )■ Figure 3.19: Adding additional points to the graph of y There are an infinite number of points that satisfy the equation y = x 2 — 7. In Figure 3.19, we've plotted only 17 points that satisfy the equation y = x 2 — 7. However, the collection of points in Figure 3.19 suggest that if we were to plot the remainder of the points that satisfy the equation y = x 2 — 7, the result would be the curve (called a parabola) shown in Figure 3.20. 164 CHAPTER 3. INTRODUCTION TO GRAPHING 10 y = x 2 - 7 -► x 10 -10- m -10- Figure 3.20: The graph of y = x 2 — 7 is a curve called a parabola. □ 3.1. GRAPHING EQUATIONS BY HAND 165 **>**■**> Exercises ■** * «•* In Exercises 1-6, set up a Cartesian Coordinate system on a sheet of graph paper, then plot the given points. 1. A(2,-4), B(-4,-3), and C(-3,2) 2. 4(3,-4), 5(-3,-2), and C(-4,4) 3. 4(3,-2), 5(-3,-4), and C(-2, 3) 4. 4(3,-4), B(-3,-4), and C(-4,2) 5. A(-2,2), 5(-2,-3), and C(2, -2) 6. A(-3,2), 5(2,-4), and C(-3, -4) In Exercises 7-10, identify the coordinates of point P in each of the given coordinate systems. 7. 9. y 10. 166 CHAPTER 3. INTRODUCTION TO GRAPHING 11. Which of the points (8, 36), and (—2, —12) is a solution y = 5x — 5? 12. Which of the points (0, (1,-14), and (-6,36) is a equation y = — 7x — 7? 13. Which of the points (- (-6,39), and (2,-3) is a equation y = — 5x + 6? 14. Which of the points (7, - (-6,19), and (5,-11) is a equation y = — 3a; + 3? (5,20), (4,13), of the equation -9), (6,-46), solution of the -9,49), (1,1), solution of the -15), (-9,30), solution of the 15. Which of the points (1,12), (7,395), (—7,398), and (0,1) is a solution of the equation y = 8x 2 + 3? 16. Which of the points (-7,154), (7,153), (—2,21), and (2,16) is a solution of the equation y = 3a; 2 + 6? 17. Which of the points (8,400), (5,158), (0, 3), and (2, 29) is a solution of the equa- tion y = 6a; 2 + 2ai? 18. Which of the points (-6, -114), (6, -29), (—7,-149), and (4,-1) is a solution of the equation y = —2a; 2 + 7a;? In Exercises 19-26, complete the table of points that satisfy the given equation. Set up a coordinate system on a sheet of graph paper, plot the points from the completed table, then draw the graph of the given equation. 19. X y = 2x + 1 (x,y) -3 -2 -1 1 2 3 21. X y=\x-5\ (x,y) 2 3 4 5 6 7 8 20. X y = x — 4 (x,y) -3 -2 -1 1 2 3 22. X y=\x + 2\ (x,y) -5 -4 -3 -2 -1 1 3.1. GRAPHING EQUATIONS BY HAND 167 23. 24. X y = {x + l) 2 {x,y) -4 -3 -2 -1 1 2 25. X 2/=(z + 5) 2 (z,y) -8 -7 -6 -5 -4 -3 -2 26. X y = —x — 5 (a, 3/) -3 -2 -1 1 2 3 a; j/= -2a; + 3 (x,y) -3 -2 -1 1 2 3 27. Use a graphing calculator to complete a table of points satisfying the equation y = x 2 — 6x + 5. Use integer values of x, starting at and ending at 6. After completing the table, set up a coordinate system on a sheet of graph paper. Label and scale each axis, then plot the points in your table. Finally, use the plotted points as evidence to draw the graph of 29. Use a graphing calculator to complete a table of points satisfying the equation y = — x 2 + 2x + 3. Use integer values of X, starting at —2 and ending at 4. Af- ter completing the table, set up a coor- dinate system on a sheet of graph paper. Label and scale each axis, then plot the points in your table. Finally, use the plot- ted points as evidence to draw the graph y 6x + 5. of y 2x + 3. 28. Use a graphing calculator to complete a table of points satisfying the equation y = x 2 — 2x — 3. Use integer values of x, starting at —2 and ending at 4. Af- ter completing the table, set up a coor- dinate system on a sheet of graph paper. Label and scale each axis, then plot the points in your table. Finally, use the plot- ted points as evidence to draw the graph 30. Use a graphing calculator to complete a table of points satisfying the equation y = —x 2 — 2x + 8. Use integer values of x, starting at —5 and ending at 3. Af- ter completing the table, set up a coor- dinate system on a sheet of graph paper. Label and scale each axis, then plot the points in your table. Finally, use the plot- ted points as evidence to draw the graph of y 2a; -3. of y 2x 168 CHAPTER 3. INTRODUCTION TO GRAPHING 31. Use a graphing calculator to complete a table of points satisfying the equation y = x 3 — 4a; 2 — 7x + 10. Use integer val- ues of x, starting at —3 and ending at 6. After completing the table, set up a coor- dinate system on a sheet of graph paper. Label and scale each axis, then plot the points in your table. Finally, use the plot- ted points as evidence to draw the graph of y 32. 4a; 2 7x + 10. Use a graphing calculator to complete a table of points satisfying the equation y = x 3 + 3x 2 — 13a; — 15. Use integer val- ues of x, starting at —6 and ending at 4. After completing the table, set up a coor- dinate system on a sheet of graph paper. Label and scale each axis, then plot the points in your table. Finally, use the plot- ted points as evidence to draw the graph of y 3a; 2 - 13a; - 15. 33. Use a graphing calculator to complete a table of points satisfying the equation y = —x 3 + Ax 2 + 7x — 10. Use integer val- ues of x, starting at —3 and ending at 6. After completing the table, set up a coor- dinate system on a sheet of graph paper. Label and scale each axis, then plot the points in your table. Finally, use the plot- ted points as evidence to draw the graph of y = -x 3 + Ax 2 + 7x- 10. 34. Use a graphing calculator to complete a table of points satisfying the equation y = —x 3 + x 2 + 12x. Use integer values of x, starting at —4 and ending at 5. Af- ter completing the table, set up a coor- dinate system on a sheet of graph paper. Label and scale each axis, then plot the points in your table. Finally, use the plot- ted points as evidence to draw the graph of y = -x 3 + x 2 + Ylx. 35. Use a graphing calculator to complete a table of points satisfying the equation y = \/x + 5. Use integer values of x, starting at —5 and ending at 10. Round your y-values to the nearest tenth. Af- ter completing the table, set up a coor- dinate system on a sheet of graph paper. Label and scale each axis, then plot the points in your table. Finally, use the plot- ted points as evidence to draw the graph of y = \/x + 5. 36. Use a graphing calculator to complete a table of points satisfying the equation y = \/4 — x. Use integer values of x, starting at —10 and ending at 4. Round your y-values to the nearest tenth. Af- ter completing the table, set up a coor- dinate system on a sheet of graph paper. Label and scale each axis, then plot the points in your table. Finally, use the plot- ted points as evidence to draw the graph of y = i/4 — x. j*- j*- j*- Answers •*$ •** ■*$ 1. 5 c(- 3,2) 5 r B(-4, - 3) L 5 A % -] ) C(-2,3) £(-3,-4). A(3, -2) 3. 1 . GRAPHING EQUATIONS BY HAND 169 23. 5^ J l(-2,L'l 5 5 , C*(2,'-2) B{ -2, -3)" 5- 7. (-4,3) 9. (3,-4) 11. (5,20) 13. (1,1) 15. (7,395) 17. (8,400) 19. -10 10 y y = 2x + 1 -10 10 21. L( J/ = , 11 i r y = \x- 5| 25. 27. y=(x + l) 2 -10 -10 104 § -io4 10 L( I / r * 1( ) lu ( ' y = —x — 5 § -y = x — dx + 5 10 170 CHAPTER 3. INTRODUCTION TO GRAPHING 29. -io- 33. 10 I 10 -10 y = -x 2 + 2x + 3 y 50 + -50 r- 10 -y = -x s + Ax 2 + 7x - 10 ltd 31. -io- 35. -50 A -50 ^! -y = X 3 - ix 2 - 7x + 10 10 LC V — , ' 10 lu ( <■ 3.2. THE GRAPHING CALCULATOR 171 3.2 The Graphing Calculator It's time to learn how to use a graphing calculator to sketch the graph of an equation. We will use the TI-84 graphing calculator in this section, but the skills we introduce will work equally well on the ancient TI-82 and the less ancient TI-83 graphing calculators. In this introduction to the graphing calculator, you will need to use several keys on the upper half of the graphing calculator (see Figure 3.21). The up- and-down and left-and-right arrow keys are located in the upper right corner of Figure 3.21. These are used for moving the cursor on the calculator view screen and various menus. Immediately below these arrow keys is the CLEAR button, used to clear the view screen and equations in the Y= menu. s-wpldi n »»„ n r0 n*u,j n caic f* o o o yM '-^ "N" HIT ^— ^^^ ^^^ ^^^ Figure 3.21: Top half of calculator. It is not uncommon that people share their graphing calculators with friends who make changes to the settings in the calculator. Let's take a moment to make sure we have some common settings on our calculators. Locate and push the MODE in the first row of Figure 3.21. This will open the window shown in Figure 3.22. Make sure that the mode settings on your calculator are identical to the ones shown in Figure 3.22. If not, use the up-and- down arrow keys to move to the non-matching line item. This should place a blinking cursor over the first item on the line. Press the ENTER button on the lower- right corner of your calculator to make the selection permanent. Once you've completed your changes, press 2ND MODE again to quit the MODE menu. Next, note the buttons across the first row of the calculator, located imme- diately below the view screen. TRACE GRAPH Above each button are one or more commands located on the calculator's case. Press the 2ND key to access a command having the same color as the 2ND key. Press the ALPHA key to access a command having the same color as the ALPHA key. The QUIT command is located above the MODE button on the calculator case and is used to exit the current menu. 172 CHAPTER 3. INTRODUCTION TO GRAPHING Just above the Y= button are the words STAT PLOT, printed in the same color as the 2ND button located near the top left of the calculator (see Figure 3.21). Press the 2ND button, then the Y= button. This opens the stat plot menu shown in Figure 3.23. FLOAT RhDIhTi nine cannECTED SEQUERTIAL REAL sci Eni] 0i234£fi7B9 DEGREE PAR PDL SEQ DDT SIHUL HORIZ G-T +TIEKT -I- Figure 3.22: Settings in the MODE window. afflfldcqp yHF , To€i..j!!rff L-1L1 L2 d 2:Plot2. ..Off LHL1 L2 a 3:Plot3. ..Off LHL1 L2 d 4-1-PlotsOff Figure 3.23: The STAT PLOT menu. You Try It! Use the graphing calculator to sketch the graph of y = —x + 3. We need all of the stat plots to be "off." If any of the three stat plots are "on," select 4:PlotsOff (press the number 4 on your keyboard), then press the ENTER key on the lower right corner of your calculator. That's it! Your calculator should now be ready for the upcoming exercises. EXAMPLE 1. Use the graphing calculator to sketch the graph of y 1. Solution: Recall that we drew the graph of y = x + 1 by hand in Example 1 of Section 3.1 (see Figure 3.9). In this example, we will use the graphing cal- culator to produce the same result. Press the Y= button. The window shown in Figure 3.24 appears. If any equations appear in the Y= menu of Figure 3.24, use the up-and-down arrow keys and the CLEAR button (located below the up-and-down and left-and- right arrow keys) to delete them. Ploti Plots Plot3 Wi = Wz = ■^3 = \Yh = ■^Ye = ^Vg = "^7 = Figure 3.24: Press the Y= button to open the Y= menu. Ploti Plots Plots WlBX+1 sV2 = \Vl = \Yh = ■^Ye = ^Vg = ^7 = Figure 3.25: Enter y Yl=. 1 in Next, move the cursor to Yl = , then enter the equation y = x + 1 in Yl with the following button keystrokes. The result is shown in Figure 3.25. 3.2. THE GRAPHING CALCULATOR 173 □ ENTER Next, select the ZOOM key on the top row of the calculator. This will open the window shown in Figure 3.26. From the ZOOM menu, select 6:ZStandard. There are two ways to make this selection: 1. Use the down-arrow key to move downward in the ZOOM menu until 6:ZStandard is highlighted, then press the ENTER key. 2. A quicker alternative is to simply press the number 6 on the calculator keyboard to select 6:ZStandard. The resulting graph of y = x + 1 is shown in Figure 3.27. Note that the result is identical to the graph of y = x + 1 drawn by hand in Example 3 of Section 3.1 (see Figure 3.9). MEMORY ox :Zoon In 3:Zoon Out 4:ZDecinal 5:ZS=iuare 6:ZStandard 7-i-ZTrig Figure 3.26: Select 6:ZStandard from the ZOOM menu. Figure 3.27: The graph of y = x- is a line. Answer: □ EXAMPLE 2. Use the graphing calculator to sketch the graph of y -7. Solution: Recall that we drew the graph of y = x 2 — 7 by hand in Example 4 of Section 3.1 (see Figure 3.20). In this example, we will use the graphing calculator to produce the same result. Press the Y= button and enter the equation y Figure 3.28) with the following keystrokes: 7 into Yl (see X,T,9,n mam ENTER You Try It! Use the graphing calculator to sketch the graph of y = -x 2 + 4. The caret (A) symbol (see Figure 3.21) is located in the last column of buttons on the calculator, just underneath the CLEAR button, and means "raised to." The caret button is used for entering exponents. For example, x 2 is entered as XA2, is entered as XA3, and so on. 174 CHAPTER 3. INTRODUCTION TO GRAPHING Answer: Press the ZOOM button, then select 6:ZStandard from the ZOOM menu to produce the graph of y = x 2 — 7 shown in Figure 3.29. Note that the result in Figure 3.29 agrees with the graph of y = x 2 — 7 we drew by hand in Example 4 of Section 3.1 (see Figure 3.20). Ploti Plots Plots ^13X^2-7 ■^£=1 \Vi = \Yh = ■^Ye = ^Vfi = \Y? = Figure 3.28: Enter y Yl. 7 in Figure 3.29: The graph of y is a parabola. □ Reproducing Calculator Results on Homework Paper In this section we delineate recommendations and requirements when repro- ducing graphing calculator results on your homework paper. Consider again the final result of Example 2 shown in Figure 3.29. To determine the scale at each end of each axis, press the WINDOW button on the topmost row of buttons on your calculator. The WINDOW settings for Figure 3.29 are shown in Figure 3.30. Figure 3.31 presents a visual explanation of each of the entries Xmin, Xmax, Ymin, and Ymax. WINDOW Xnun=-10 Xnax=l@ Xscl=l Wiin=-10 Wiax=l@ Vscl=l ■i-Xres=l Figure 3.30: The WINDOW pa- rameters. y 10 f Ymax Xmin -10 -10 Xmax x 10 Ymin Figure 3.31: Xmin, Xmax, Ymin, and Ymax contain the scale at the end of each axis. 3.2. THE GRAPHING CALCULATOR 175 Xmin and Xmax indicate the scale at the left- and right-hand ends of the a;-axis, respectively, while Ymin and Ymax indicate the scale at the bottom- and top-ends of the y-axis. Finally, as we shall see shortly, Xscl and Yscl control the spacing between tick marks on the x- and j/-axes, respectively. When reproducing the graph in your calculator viewing window on your homework paper, follow the Calculator Submission Guidelines. Calculator Submission Guidelines. 1. All lines (including the horizontal and vertical axes) should be drawn with a ruler. All curves should be drawn freehand. 2. Set up a coordinate system on your homework paper that mimics closely the coordinate system in your calculator's view screen. Label your axes (usually with x and y). 3. Indicate the scale at each end of each axis. Use Xmin, Xmax, Ymin, and Ymax in the WINDOW menu for this purpose. 4. Copy the graph from your calculator's viewing window onto your coordi- nate system. Label the graph with its equation. For example, to report the results of Figure 3.29, draw the axes with a ruler, label the horizontal axis with x, the vertical axis with y, then place the values of Xmin, Xmax, Ymin, and Ymax at the appropriate end of each axis (see Figure 3.32). Figure 3.32: Reporting the graph of y = x 2 — 7 on your homework paper. Because the graph is a curve, make a freehand copy that closely mimics the graph shown in Figure 3.29, then label it with its equation, as shown in 176 CHAPTER 3. INTRODUCTION TO GRAPHING You Try It! Approximate the point of intersection of the graphs of y = x — 4 and y = 5 — x using the TRACE button. Figure 3.32. Adjusting the Viewing Window Note that Figure 3.31 gives us some sense of the meaning of the "standard viewing window" produced by selecting 6:ZStandard from the ZOOM menu. Each time you select 6:ZStandard from the ZOOM menu, Xmin is set to —10, Xmax to 10, and Xscl to 1 (distance between the tick marks on the x-axis). Similarly, Ymin is set to —10, Ymax to 10, and Yscl to 1 (distance between the tick marks on the y-axis). You can, however, override these settings as we shall see in the next example. EXAMPLE 3. Sketch the graphs of the following equations on the same viewing screen. 5 , 2 y = —x — 3 and y = —x + 4 y 4 y 3 Adjust the viewing window so that the point at which the graphs intersect (where the graphs cross one another) is visible in the viewing window, then use the TRACE button to approximate the coordinates of the point of intersection. Solution: We must first decide on the proper syntax to use when entering the equations. In the case of the first equation, note that y is pronounced "y equals five-fourths x minus three," and means "y equals five- fourths times x minus three." Enter this equation into Yl in the Y= menu as 5/4*X— 3 (see Figure 3.33), using the button keystrokes: 0H0 The division and times buttons are located in the rightmost column of the calculuator. Similarly, enter the second equation into Y2 as 2/3*X+4 using the button keystrokes: 0B0 A, T, 9, n □ Select 6:ZStandard in the ZOOM menu to produce the lines in Figure 3.34. When we examine the resulting graphs in Figure 3.34, it appears that their point of intersection will occur off the screen above and to the right of the upper right-corner of the current viewing window. With this thought in mind, let's extend the a;-axis to the right by increasing the value of Xmax to 20. Further, let's extend the y-axis upward by increasing the value of Ymin to 20. Press the WINDOW button on the top row of the calculator, then make the adjustments shown in Figure 3.35. 3.2. THE GRAPHING CALCULATOR 177 Ploti Plots Plots WiB5^4*tf-3 ^eB2^3*X+4 ^7 = Figure 3.33: Enter equations. Figure 3.34: The graphs. WINDOW Xnun=-10 Xnax=2@ Xscl=l Vmin=-10 Wiax=20 Vscl=l ■i-Xres=l . . . . .^^. . . Figure 3.35: Adjusting the viewing window. Figure 3.36: The point of intersec- tion is visible in the new viewing window. If we select 6:ZStandard from the ZOOM menu, our changes to the WIN- DOW parameters will be discarded and the viewing window will be returned to the "standard viewing window." If we want to keep our changes to the WIN- DOW parameters, the correct approach at this point is to push the GRAPH button on the top row of the calculator. The resulting graph is shown in Figure 3.36. Note that the point of intersection of the two lines is now visible in the viewing window. The image in Figure 3.36 is ready for recording onto your homework. How- ever, we think we would have a better picture if we made a couple more changes: 1. It would be nicer if the point of intersection were more centered in the viewing window. 2. There are far too many tick marks. With these thoughts in mind, make the changes to the WINDOW parame- ters shown in Figure 3.37, then push the GRAPH button to produce the image in Figure 3.38. 178 CHAPTER 3. INTRODUCTION TO GRAPHING WINDOW Xnin=l5 Xnax=25 Xscl=5 Ynun= - 5 Wiax=25 Vscl=5 ■i-Xres=l Figure 3.37: A final adjustment of the viewing window. Figure 3.38: A centered point of in- tersection and fewer tick marks. The TRACE button is only capable of providing a very rough approximation of the point of intersection. In Chapter 4, we'll introduce the intersection utility on the CALC menu, which will report a much more accurate result. Answer: (4.68,0.68) Y1=K-H"-v K=1.fiH0HEli V=.fiH0HE10fi Finally, push the TRACE button on the top row of buttons, then use the left- and right-arrow keys to move the cursor atop the point of intersection. An approximation of the coordinates of the point of intersection are reported at the bottom of the viewing window (see Figure 3.39). V1=E/H*K-3 K=ii.91HB9H Y=ii.H93fii7 Figure 3.39: Approximate coordi- nates of the point of intersection re- ported by the TRACE button. Figure 3.40: Reporting the answer on your homework. In reporting the answer on our homework paper, note that we followed the Calculator Submission Guidelines on page 175. □ 3.2. THE GRAPHING CALCULATOR 179 **. i*. **. Exercises ■*$*$-*$ In Exercises 1-16, enter the given equation in the Y= menu of your calcuator, then select 6:ZStandard from the ZOOM menu to produce its graph. Follow the Calculator Submission Guidelines outlined in the subsection Reproducing Calculator Results on Homework Paper when submitting your result on your homework paper. 1. y = 2x + 3 2. y = 3- 2x 3. y = -x - 4 y 2 2 4. y = --x + 2 y 3 5. y = 9-x 2 6. y = x 2 -4 7. y = -x 2 - 3 y 2 8. h = 4 x 2 y 3 9. y = \fx 10. y = y = y = ^-x 11. Vx + 3 12. \/5 — x 13. y = |*| 14. y = -N 15. y = |» + 2| 16. y = -|a: -3 In Exercises 17-22, the graph of each given equation is a parabola, having a "u-shape" similar to the graph in Figure 3.32. Some of the parabolas "open up" and some "open down." Adjust Ymin and/or Ymax in the WINDOW menu so that the "turning point" of the parabola, called its vertex, is visible in the viewing screen of your graphing calculator. Follow the Calculator Submission Guidelines outlined in the subsection Reproducing Calculator Results on Homework Paper when submitting your result on your homework paper. 17.2/ = x 2 + x - 20 18. y = -x 2 - 2a; + 24 19. y = -x 2 +4x + 21 20. y = x 2 + 6x - 16 21. y = 2x 2 - 13x-24 22. y = -3x 2 - 19a; + 14 180 CHAPTER 3. INTRODUCTION TO GRAPHING In Exercises 23-28, sketch the given lines, then adjust the WINDOW parameters so that the point of intersection of the two lines is visible in the viewing window. Use the TRACE key to estimate the point of intersection. Follow the Calculator Submission Guidelines outlined in the subsection Reproducing Calculator Results on Homework Paper when submitting your result on your homework paper. Be sure to also include your estimate of the point of intersection in your sketch. 23. y = 2x + 1 and y = x + ; 24. y = 4 — 2x and y = — 6 - 1 2 25. y = —x and y = —x — 3 y 2 J 3 26. y = 2x + 5 and y = x — 27. y 28. y = 5 — x and y 1 2 = -3- 2x 1 , 4 = 1 x and y = —4 x 2 5 J*. t»- ;*■ Answers •*;■*; •*! y = 2t + 3 10 . -10 ' —10 \ 10 1 2/ = 9 - x 2 3. y=2 x 12 o . 10' . -10 -10 / 10 ■ 3.2. THE GRAPHING CALCULATOR 181 10 -10 -10 y = v^ 10 15. y y-\x + 2\ 10 - -10 -10 10 11. 17. -10 y — \Jx + 3 2/ y = x 2 + x - 20 30 . -10 -30 / 10 1 13. 19. V = 1*1 j/ = -x 2 +4x + 21 182 CHAPTER 3. INTRODUCTION TO GRAPHING 21. 25. y y = 2x 2 - 13a; - 24 . 50 , -10 -50 / 10 y = 9 x (18.08,9.04) 35 23. y = x + 27. y = 2x + l y — 5 — x y = -3-2x 3.3. RATES AND SLOPE 183 3.3 Rates and Slope Let's open this section with an application of the concept of rate. EXAMPLE 1. An object is dropped from rest, then begins to pick up speed at a constant rate of 10 meters per second every second (10 (m/s)/s or 10 m/s 2 ). Sketch the graph of the speed of the object versus time. Solution: In this example, the speed of the object depends on the time. This makes the speed the dependent variable and time the independent variable. Independent versus dependent. It is traditional to place the independent variable on the horizontal axis and the dependent variable on the vertical axis. You Try It! Starting from rest, an automobile picks up speed at a constant rate of 5 miles per hour every second (5 (mi/hr)/s). Sketch the graph of the speed of the object versus time. Following this guideline, we place the time on the horizontal axis and the speed on the vertical axis. In Figure 3.41, note that we've labeled each axis with the dependent and independent variables (v and t), and we've included the units (m/s and s) in our labels. Next, we need to scale each axis. In determining a scale for each axis, keep two thoughts in mind: 1. Pick a scale that makes it convenient to plot the given data. 2. Pick a scale that allows all of the given data to fit on the graph. In this example, we want a scale that makes it convenient to show that the speed is increasing at a rate of 10 meters per second (10 m/s) every second (s). One possible approach is to make each tick mark on the horizontal axis equal to 1 s and each tick mark on the vertical axis equal to 10 m/s. v(m/s) 100 90 80 70 CO 50 40 30 20 10 v(m/s) 100 90 80 70 (30 50 40 30 20 10 i(s) 0123456789 10 Figure 3.41: Label and scale each axis. Include units with labels. ' I I I I I I I I I t <> — — ii — ii ii ii ii ii ii — ii n — I — I — I — I — I — I — I — I — I — L> 0123456789 10 t(s) Figure 3.42: Start at (0,0), then continuously move 1 right and 10 up. 184 CHAPTER 3. INTRODUCTION TO GRAPHING Next, at time t = s, the speed is v = m/s. This is the point (t, v) = (0, 0) plotted in Figure 3.42. Secondly, the rate at which the speed is increasing is 10 m/s per second. This means that every time you move 1 second to the right, the speed increases by 10 m/s. In Figure 3.42, start at (0,0), then move 1 s to the right and 10 m/s up. This places you at the point (1, 10), which says that after 1 second, the speed of the particle is 10 m/s. Continue in this manner, continuously moving 1 s to the right and 10 m/s upward. This produces the sequence of points shown in Figure 3.42. Note that this constant rate of 10 (m/s)/s forces the graph of the speed versus time to be a line, as depicted in Figure 3.43. Answer: u(mi/hr) i>(m/s) 100 0123456789 10 Figure 3.43: The constant rate forces the graph to be a line. □ Measuring the Change in a Variable To calculate the change in some quantity, we take a difference. For example, suppose that the temperature in the morning is 40° F, then in the afternoon the temperature measures 60° F (F stands for Fahrenheit temperature). Then the change in temperature is found by taking a difference. Change in temperature = Afternoon temperature — Morning temperature = 60° F - 40° F = 20° F Therefore, there was a twenty degree increase in temperature from morning to afternoon. 3.3. RATES AND SLOPE 185 Now, suppose that the evening temperature measures 50° F. To calculate the change in temperature from the afternoon to the evening, we again subtract. Change in temperature = Evening temperature — Afternoon temperature = 50°F-60°F = -10° F There was a ten degree decrease in temperature from afternoon to evening. Calculating the Change in a Quantity. To calculate the change in a quantity, subtract the earlier measurement from the later measurement. Let T represent the temperature. Mathematicians like to use the symbolism AT to represent the change in temperature. For the change in temperature from morning to afternoon, we would write AT = 20° F. For the afternoon to evening change, we would write AT = —10° F. Mathematicians and scientists make frequent use of the Greek alphabet, the first few letters of which are: a, j3, 7, d, . . . (Greek alphabet, lower case) A, B,T,A,... (Greek alphabet, upper case) a, b,c,d, . . . (English alphabet) Thus, the Greek letter A, the upper case form of 5, correlates with the letter 'rf' in the English alphabet. Why did mathematicians make this choice of letter to represent the change in a quantity? Because to find the change in a quantity, we take a difference, and the word "difference" starts with the letter 'd.' Thus, AT is also pronounced "the difference in T." Important Pronunciations. Two ways to pronounce the symbolism AT. 1. AT is pronounced "the change in T." 2. AT is also pronounced "the difference in T." Slope as Rate Here is the definition of the slope of a line. Slope. The slope of a line is the rate at which the dependent variable is chang- ing with respect to the independent variable. For example, if the dependent 186 CHAPTER 3. INTRODUCTION TO GRAPHING variable is y and the independent variable is x, then the slope of the line is: Ay Slope Ax You Try It! Starting from rest, an automobile picks up speed at a constant rate of 5 miles per hour every second (5 (mi/hr)/s). The constant rate forces the graph of the speed of the object versus time to be a line. Calculate the slope of this line. EXAMPLE 2. In Example 1, an object released from rest saw that its speed increased at a constant rate of 10 meters per second per second (10 (m/s)/s or 10m/s 2 ). This constant rate forced the graph of the speed versus time to be a line, shown in Figure 3.43. Calculate the slope of this line. Solution: Start by selecting two points -P(2,20) and Q(8,80) on the line, as shown in Figure 3.44. To find the slope of this line, the definition requires that we find the rate at which the dependent variable v changes with respect to the independent variable t. That is, the slope is the change in v divided by the change in t. In symbols: Slope = — — 1 At w(m/s) 100 90 80 70 60 50 40 30 20 10 . 1-- )(8 ■ a - 1 J p (•>■ 9f ) 12 3 4 5 6 7 9 10 t( S ) u(m/s) 100 90 80 70 60 50 40 30 20 10 . Q )(7, to) p (3 3r ) 0123456789 10 too Figure 3.45: The slope does not de- pend on the points we select on the line. Figure 3.44: Pick two points to compute the slope. Now, as we move from point P(2, 20) to point Q(8, 80), the speed changes from 20 m/s to 80 m/s. Thus, the change in the speed is: Aw = 80 m/s- 20 m/s = 60 m/s 3.3. RATES AND SLOPE 187 Similarly, as we move from point P(2,20) to point Q(8, 80), the time changes from 2 seconds to 8 seconds. Thus, the change in time is: A^ = 8s- 2s = 6s Now that we have both the change in the dependent and independent variables, we can calculate the slope. Slope = — — At 60m/s ~ 6s = 10^ s Therefore, the slope of the line is 10 meters per second per second (10 (m/s)/s or 10m/s 2 ). The slope of a line does not depend upon the points you select. Let's try the slope calculation again, using two different points and a more compact presentation of the required calculations. Pick points P(3,30) and Q(7, 70) as shown in Figure 3.45. Using these two new points, the slope is the rate at which the dependent variable v changes with respect to the independent variable t. Ql AV Slope = — — At 70 m/s - - 30 m/s 7s- -3s 40 m/s 4s m/s = 10 — s Again, the slope of the line is 10 (m/s)/s. Answer: 5 (mi/hr)/s □ Example 2 points out the following fact. Slope is independent of the selected points. It does not matter which two points you pick on the line to calculate its slope. The next example demonstrates that the slope is also independent of the order of subtraction. 188 CHAPTER 3. INTRODUCTION TO GRAPHING You Try It! Compute the slope of the line passing through the points P(-3, 1) and <5(2, 4). EXAMPLE 3. Compute the slope of the line passing through the points P(-l,-2) and (3(3,3). Solution: First, sketch the line passing through the points P(— 1, — 2) and Q(3, 3) (see Figure 3.46). Warning! If you are not consistent in the direction you subtract, you will not get the correct answer for the slope. For example: (-2) -1 In this case, we subtracted the y-coordinate of point P(-l,-2) from the y-coordinate of point Q(3, 3), but then we changed horses in midstream, subtracting the x-coordinate of point (5(3, 3) from the cc-coordinate of point P(-l, -2). Note that we get the negative of the correct answer. Figure 3.46: Computing the slope of the line passing through the points P(-l,-2) and Q(3,3). To calculate the slope of the line through the points P(— 1, —2) and <5(3, 3), we must calculate the change in both the independent and dependent variables. We'll do this in two different ways. Subtract the coordinates of point P(— 1,— 2) from the coordinates of point (5(3, 3). Slope = _ Ay Air 3-(- -2) 3-(- -1) 5 " 4 Subtract the coordinates of point Q(3,3) from the coordinates of point P(-l,-2). Slope = — — Air -1-3 -5 Answer: 3/5 _ 5 ~ 4 Note that regardless of the direction of subtraction, the slope is 5/4. □ 3.3. RATES AND SLOPE 189 Example 3 demonstrates the following fact. The direction of subtraction does not matter. When calculating the slope of a line through two points P and Q, it does not matter which way you subtract, provided you remain consistent in your choice of direction. The Steepness of a Line We need to examine whether our definition of slope matches certain expecta- tions. Slope and steepness of a line. The slope of a line is a number that tells us how steep the line is. If slope is a number that measures the steepness of a line, then one would expect that a steeper line would have a larger slope. EXAMPLE 4. Graph two lines, the first passing through the points P(— 3, — 2) and Q(3,2) and the second through the points R(— 1,— 3) and 5(1,3). Calcu- late the slope of each line and compare the results. Solution: The graphs of the two lines through the given points are shown, the first in Figure 3.47 and the second in Figure 3.48. Note that the line in Figure 3.47 is less steep than the line in Figure 3.48. V S (1. 3) -5 5 L -1. R{- *) 5- Figure 3.47: This line is less steep than the line on the right. Figure 3.48: This line is than the line on the left. steeper Remember, the slope of the line is the rate at which the dependent variable is changing with respect to the independent variable. In both Figure 3.47 and Figure 3.48, the dependent variable is y and the independent variable is x. You Try It! Compute the slope of the line passing through the points P(— 2, —3) and Q(2,5). Then compute the slope of the line passing through the points 7?(-2,-l) and 5(5,3), and compare the two slopes. Which line is steeper? 190 CHAPTER 3. INTRODUCTION TO GRAPHING Answer: The first line has slope 2, and the second line has slope 4/7. The first line is steeper. You Try It! Compute the slope of the line passing through the points P(— 3, 3) and Q(3,— 5). Then compute the slope of the line passing through the points R(— 4, 1) and 5(4, —3), and compare the two slopes. Which line is steeper? Subtract the coordinates of point P(— 3,— 2) from the coordinates of point Q(3,2). Subtract the coordinates of the point R(—l,—3) from the point 5(1,3). Slope of first line = — — Ay Ax 2 -(-2) Ay Slope of second line = — — Ax (-3) (-3) 1 G 2 3 (-1) Note that both lines go uphill and both have positive slopes. Also, note that the slope of the second line is greater than the slope of the first line. This is consistent with the fact that the second line is steeper than the first. □ In Example 4, both lines slanted uphill and both had positive slopes, the steeper of the two lines having the larger slope. Let's now look at two lines that slant downhill. EXAMPLE 5. Graph two lines, the first passing through the points P(— 3, 1) and (5(3,-1) and the second through the points R(— 2,4) and 5(2,-4). Cal- culate the slope of each line and compare the results. Solution: The graphs of the two lines through the given points are shown, the first in Figure 3.49 and the second in Figure 3.50. Note that the line in Figure 3.49 goes downhill less quickly than the line in Figure 3.50. Remember, the slope of the line is the rate at which the dependent variable is changing with respect to the independent variable. In both Figure 3.49 and Figure 3.50, the dependent variable is y and the independent variable is x. Subtract the coordinates of point P(— 3, 1) from the coordinates of point Q(3, —1). Slope of first line = Ax -1- 1 3-(- -3) -2 6 1 3 Subtract the coordinates of point R(— 2,4) from the coordinates of point 5(2,-4). Ay Slope of second line = — — Ax -4-4 r F2) 3.3. RATES AND SLOPE 191 5^ i P (" 3. i) 5 5 Q (3 1) 5 Figure 3.49: This line goes down- hill more slowly than the line on the right. ■J / R(- 2. 1) -5 5 (2. -4) S ■ r >; ■ Figure 3.50: This line goes downhill more quickly than the line on the left. Note that both lines go downhill and both have negative slopes. Also, note that the magnitude (absolute value) of the slope of the second line is greater than the magnitude of the slope of the first line. This is consistent with the fact that the second line moves downhill more quickly than the first. Answer: The first line has slope —4/3, and the second line has slope —1/2. The first line is steeper. □ What about the slopes of vertical and horizontal lines? EXAMPLE 6. Calculate the slopes of the vertical and horizontal lines passing through the point (2,3). Solution: First draw a sketch of the vertical and horizontal lines passing through the point (2,3). Next, select a second point on each line as shown in Figures 3.51 and 3.52. You Try It! Calculate the slopes of the vertical and horizontal lines passing through the point (-4,1). 5t ■P(-2,3)— Q(2,3) Figure 3.51: A horizontal line through (2,3). '/ <-S(2,3) *■ x fl(2,-3) Figure 3.52: A vertical line through (2,3). 192 CHAPTER 3. INTRODUCTION TO GRAPHING The slopes of the horizontal and vertical lines are calculated as follows. Subtract the coordinates of the point P(— 2,3) from the coordi- nates of the point Q(2, 3). Slope of horizontal line Ay Ax _3-3 2~ 4 (-2) Subtract the coordinates of the point (2,-3) from the coordinates of the point 5(2,3). Slope of vertical line Ay Ace 3 ~(-3) 2-2 6 undefined Answer: The slope of the vertical line is undefined. The slope of the second line is 0. Thus, the slope of the horizontal line is zero, which makes sense be- cause a horizontal line neither goes uphill nor downhill. Division by zero is undefined. Hence, the slope of a vertical line is undefined. Again, this makes sense because as uphill lines get steeper and steeper, their slopes increase without bound. □ The Geometry of the Slope of a Line We begin our geometrical discussion of the slope of a line with an example, calculating the slope of a line passing through the points P(2,3) and Q(8, 8). Before we begin we'll first calculate the change in y and the change in x by sub- tracting the coordinates of point -P(2, 3) from the coordinates of point Slope Ay Ax 8-3 _ 5 ~~ 6 Thus, the slope of the line through the points P(2, 3) and <5(8, 8) is 5/6. To use a geometric approach to finding the slope of the line, first draw the line through the points P(2,3) and (3(8,8) (see Figure 3.53). Next, draw a right triangle with sides parallel to the horizontal and vertical axes, using the points P(2,3) and Q(8,8) as vertices. As you move from point P to point R in Figure 3.53, note that the change in x is Ax = 6 (count the tick marks 1 ). 1 When counting tick marks, make sure you know the amount each tick mark represents. For example, if each tick mark represents two units, and you count six tick marks when evaluating the change in x, then Ax = 12. 3.3. RATES AND SLOPE 193 As you then move from point R to point Q, the change in y is Ay = 5 (count the tick marks). Thus, the slope is Ay/ Ax = 5/6, precisely what we got in the previous computation. Figure 3.53: Determining the slope of the line from the graph. 10 R(6,S) Ax — b Q (8 8) Ay = r, P (2 3) -1_ 1 in Figure 3.54: Determining the slope of the line from the graph. For contrast, in Figure 3.54, we started at the point P(2, 3), then moved upward 5 units and right 6 units. However, the change in y is still Ay = 5 and the change in x is still Ax = 6 as we move from point P(2,3) to point Q(8,8). Hence, the slope is still Ay /Ax = 5/6. Consider a second example shown in Figure 3.55. Note that the line slants downhill, so we expect the slope to be a negative number. #(8,7) I in' (2 7) P \( 1 = - [ (8 3) Q R( 2- 3) L \X = fi 1 p — 1 1 Ay = -4 * x Figure 3.55: Determining the slope of the line from the graph. Figure 3.56: Determining the slope of the line from the graph. Rise over run. In Figure 3.54, we start at the point P(2,3), then "rise" 5 units, then "run" 6 units to the right. For this reason, some like to think of the slope as "rise over run." In Figure 3.55, we've drawn a right triangle with sides parallel to the hor- In this case, the "rise" is izontal and vertical axes, using the points P(2,7) and Q(8,3) as vertices. As negative, while the "run" is you move from point P to point R in Figure 3.55, the change in y is Ay = —4 positive. 194 CHAPTER 3. INTRODUCTION TO GRAPHING (count the tick marks and note that your values of y are decreasing as you move from P to R). As you move from point R to point Q, the change is x is A2: = 6 (count the tick marks and note that your values of x are increasing as you move from R to Q). Thus, the slope is Ay/ Ax = —4/6, or —2/3. Note that the slope is negative, as anticipated. In Figure 3.56, we've drawn our triangle on the opposite side of the line. In this case, as you move from point P to point R in Figure 3.56, the change in x is Ax = 6 (count the tick marks and note that your values of x are increasing as you move from P to R). As you move from point R to point Q, the change is y is Ay = — 4 (count the tick marks and note that your values of y are decreasing as you move from J? to Q). Thus, the slope is still Ay/Ax = —4/6, or —2/3. We can verify our geometrical calculations of the slope by subtracting the coordinates of the point P (2, 7) from the point Q(8, 3). Q1 Ay Slope = — — Ax 3-7 You Try It! Sketch the line passing through the point (—4, 2) with slope —1/4. 2 ~ ~3 This agrees with the calculations made in Figures 3.55 and 3.56. Let's look at a final example. EXAMPLE 7. Sketch the line passing through the point (—2, 3) with slope -2/3. Solution: The slope is —2/3, so the line must go downhill. In Figure 3.57, we start at the point P{— 2,3), move right 3 units to the point _R(1,3), then move down 2 units to the point Q(l, 1). Draw the line through the points P and Q and you are done. In Figure 3.58, we take a different approach that results in the same line. Start at the point P'(— 2,3), move downward 4 units to the point R'{— 2, — 1), then right 6 units to the point Q'(4, — 1). Draw a line through the points P' and Q' and you are done. The triangle APQR in Figure 3.57 is similar to the triangle AP'Q'R' in Figure 3.58, so their sides are proportional. Consequently, the slope of the line through points P'(-2,3) and Q'(4,-l), Slope Ay Ax -4 ir 2 ~3 3.3. RATES AND SLOPE 195 b A r = 3 fl(l,3) P{- -2, 3) 2— Q (1, 1) 5 5 -5- Figure 3.57: Start at P(-2, 3), then move right 3 and down 2. The re- sulting line has slope —2/3. .y ^4 P J (-2,3) Ay -4 R' 5 i u (-2,-1) Ax = ( ^ ■ ■ Figure 3.58: Starting at P'(-2,3) and moving down 4 and right 6 also yields a slope of —2/3. reduces to the slope of the line through the points P and Q in Figure 3.57. Answer: -5 J 1 -P(- Z O"! ) - f >(0, << 5 5 - b □ A summary of facts about the slope of a line. We present a summary of facts learned in this section. 1. The slope of a line is the rate at which the dependent variable is changing with respect to the independent variable. If y is the dependent variable and x is the independent variable, then the slope is Q1 Ay Slope =— , where Ay is the change in y (difference in y) and Aa; is the change in x (difference in x). 2. If a line has positive slope, then the line slants uphill as you "sweep your eyes from left to right." If two lines have positive slope, then the line with the larger slope rises more quickly. 3. If a line has negative slope, then the line slants downhill as you "sweep your eyes from left to right." If two lines have negative slope, then the line having the slope with the larger magnitude falls more quickly. 4. Horizontal lines have slope zero. 5. Vertical lines have undefined slope. 196 CHAPTER 3. INTRODUCTION TO GRAPHING **. **• t* Exercises •** •** •** 1. An object's initial velocity at time t = seconds is w = 10 meters per second. It then begins to pick up speed (accelerate) at a rate of 5 meters per second per second (5 m/s/s or 5m/s 2 ). a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. Include units with your labels. b) Plot the point representing the initial velocity at time t = seconds. Then plot a minimum of 5 additional points using the fact that the object is accelerating at a rate of 5 meters per second per second. c) Sketch the line representing the object's velocity versus time. d) Calculate the slope of the line. 2. An object's initial velocity at time t = seconds is v = 40 meters per second. It then begins to lose speed at a rate of 5 meters per second per second (5 m/s/s or 5m/s 2 ). a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. Include units with your labels. b) Plot the point representing the initial velocity at time t = seconds. Then plot a minimum of 5 additional points using the fact that the object is losing speed at a rate of 5 meters per second per second. c) Sketch the line representing the object's velocity versus time. d) Calculate the slope of the line. 3. David first sees his brother when the distance separating them is 90 feet. He begins to run toward his brother, decreasing the distance d between him and his brother at a constant rate of 10 feet per second (10 ft/s). a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. Include units with your labels. b) Plot the point representing David's initial distance from his brother at time t = seconds. Then plot a minimum of 5 additional points using the fact that David's distance from his brother is decreasing at a constant rate of 10 feet per second (10 ft/s). c) Sketch the line representing David's distance from his brother versus time. d) Find the slope of the line. 3.3. RATES AND SLOPE 197 4. David initially stands 20 feet from his brother when he sees his girl friend Mary in the distance. He begins to run away from his brother and towards Mary, increasing the distance d between him and his brother at a constant rate of 10 feet per second (10 ft/s). a) Set up a Cartesian Coordinate System on a sheet of graph paper. Label and scale each axis. Include units with your labels. b) Plot the point representing David's initial distance from his brother at time t = seconds. Then plot a minimum of 5 additional points using the fact that David's distance from his brother is increasing at a constant rate of 10 feet per second (10 ft/s). c) Sketch the line representing David's distance from his brother versus time. d) Find the slope of the line. In Exercises 5-14, calculate the slope of the line passing through the points P and Q. Be sure to reduce your answer to lowest terms. 5. P(9,0),Q(-9,15) 6. P(19,-17), Q(-13,19) 7. P(0,11), Q(16,-ll) 8. P(-10,-8),Q(11,19) 9. P(H,1), Q(-l,-l) 10. P(16,-15), £(-11,12) 11. P(-18,8), Q(3,-10) 12. P(9,9), Q(-6,3) 13. P(-18,10), Q(-9,7) 14. P(-7,20), Q(7,8) In Exercises 15-18, use the right triangle provided to help determine the slope of the line. Be sure to pay good attention to the scale provided on each axis when counting boxes to determine the change in y and the change in x. 15. 16. *■ x 10 . \ \ *■ X 20 198 CHAPTER 3. INTRODUCTION TO GRAPHING 17. 18. n 1 / u 5 5 -1 \ r ' o 10 10 / 5- 19. On one coordinate system, sketch each of the lines that pass through the fol- lowing pairs of points. Label each line with its slope, then explain the relationship between the slope found and the steepness of the line drawn. a) (0,0) and (1,1) b) (0,0) and (1,2) c) (0,0) and (1,3) 20. On one coordinate system, sketch each of the lines that pass through the fol- lowing pairs of points. Label each line with its slope, then explain the relationship between the slope found and the steepness of the line drawn. a) (0,0) and (1,-1) b) (0,0) and (1,-2) c) (0,0) and (1,-3) In Exercises 21-30, setup a coordinate system on graph paper, then sketch the line passing through the point P with the slope m. 21. P(-4,0), m = -3/7 22. P(-3,0), m = -3/7 23. P(-3,0), m = 3/7 24. P(-3,0), m = 3/4 25. P(-3,-3), m = 3/7 26. P(-2,3), m= -3/5 27. P(-4,3), m = -3/5 28. P(-l,-3), m = 3/4 29. P(-1,0), m= -3/4 30. P(-3,3), m= -3/4 3.3. RATES AND SLOPE 199 £*• 2*> £*■ Answers ■*$ -as >*$ v(m/s) 50 40 30 20 10 - . a 1 1 fV V -•-v 0123456789 10 d(ft) t(s) 100 90 80 70 GO 50 40 30 20 10 ♦ (( u )U, 0123456789 10 t(s) 5.-* 6 17. 19. 21. -m 3 = 3 Ay = -3 -54 P(-4,0) A. -0(3,-3) r.-ii 23. 11. = 7" Ay = _Q(4,3) 13. — 3 _P(-3,0). 15.5 2 200 CHAPTER 3. INTRODUCTION TO GRAPHING 25. Ay V -54 Ax = 1 P(-3,-3). 1-5-L-L 4(4,0) 29. 0(3,-3) 27. Ay = 5 3) i (- % 3 \ (l " Q 5 A; c = = 5 5 \ s. r. 3.4. SLOPE-INTERCEPT FORM OF A LINE 201 3.4 Slope-Intercept Form of a Line We start with the definition of the y-intercept of a line. The y-intercept. The point (0, 6) where the graph of a line crosses the y-axis is called the y-intercept of the line. We will now generate the slope-intercept formula for a line having y-intercept (0, b) and Slope = m (see Figure 3.59). Let (x, y) be an arbitrary point on the line. Slope = m *■ x Figure 3.59: Line with y-intercept at (0, b) and Slope = m. Start with the fact that the slope of the line is the rate at which the dependent variable is changing with respect to the independent variable. Slope Ay Ax Slope formula. Substitute m for the slope. To determine both the change in y and the change in x, subtract the coordinates of the point (0, b) from the point (x, y). y-b x-0 y-b Substitute m for the Slope. Ay = y — b and Ax = x — 0. Simplify. Clear fractions from the equation by multiplying both sides by the common denominator. y-b mx = y — b Multiply both sides by x. Cancel. 202 CHAPTER 3. INTRODUCTION TO GRAPHING Solve for y. mx + b = y— b + b mx + b = y Add b to both sides. Simplify. Thus, the equation of the line is y = mx + b. The Slope-Intercept form of a line. The equation of the line having y- intercept (0, b) and slope m is: y = nix + b Because this form of a line depends on knowing the slope m and the intercept (0,6), this form is called the slope-intercept form of a line. You Try It! Sketch the graph of the line having equation y -x- 2. EXAMPLE 1. Sketch the graph of the line having equation y = —x + 1. 5 Solution: Compare the equation y = |a; + 1 with the slope-intercept form y = mx + b, and note that m = 3/5 and 6=1. This means that the slope is 3/5 and the y-intercept is (0, 1). Start by plotting the y-intercept (0, 1), then move uward 3 units and right 5 units, arriving at the point (5, 4). Draw the line through the points (0, 1) and (5, 4), then label it with its equation y = |ir + 1. V 3 : ~y - 7 a ^ Ax = 5 ^ U" Aw = 3 ; ' "(5,4)- ^_ .y "(0. 1) r -1U ^^ 1U .^ S S J — 10- J- x + l Figure 3.60: Hand-drawn graph of Figure 3.61: Select 6:ZStandard from the ZOOM menu to draw the graph of y ix + 1. y lx + 1. Check: Enter the equation y 1 in Yl in the Y= menu as 3/5*X+l. Select 6:ZStandard from the ZOOM menu to produce the graph shown in 3.4. SLOPE-INTERCEPT FORM OF A LINE 203 Figure 3.61. When we compare the calculator produced graph in Figure 3.61 to the hand drawn graph in Figure 3.60, they look a bit different. This difference is due to the fact that the calculator's viewing window is wider than it is tall. On the other hand, the grid in Figure 3.60 is perfectly square. Hence, it appears that the lines increase at different angles. The graphing calculator has a feature that will cure this distortion. Press the ZOOM button, the select 5:ZSquare to produce the image in Figure 3.63. V 3 : y-K x s Ax = 5 ^ U" Au = 3 ; ;■* "(5,4)- <s_ .' (0, 1) r -10 ^^ 1U *7 / S J — 10^ J- Figure 3.63: Select 5:ZSquare from the ZOOM menu to draw the graph of y ix + 1. Figure 3.62: Hand-drawn graph of V= tX+1. When we compare the calculator image in Figure 3.63 with the hand-drawn graph in Figure 3.62, we get a better match. Answer: 4 y 10 4 / L y > -10 5 H 10 \ 1 10 -2 s ) ^ /"Q «\- \ (3, — b) LL— io- D EXAMPLE 2. Sketch the line with y-intercept (0, 2) and slope -7/3. Label the line with the slope-intercept form of its equation. Solution: Plot the y-intercept (0,2). Now use the slope —7/3. Start at (0,2), then move down seven units, followed by a three unit move to the right to the point (3,-5). Draw the line through the points (0,2) and (3,-5). (see Figure 3.64). Next, the y-intercept is (0,2), so b = 2. Further, the slope is —7/3, so m = —7/3. Substitute these numbers into the slope-intercept form of the line. y y mx 7 Slope-intercept form. Substitute: —7/3 for m, 2 for b. Therefore, the slope-intercept form of the line is y with this equation. 2. Label the line You Try It! Sketch the line with y-intercept (0, —3) and slope 5/2. Label the line with the slope-intercept form of its equation. 204 CHAPTER 3. INTRODUCTION TO GRAPHING y = - -x + 2 y yfy -J^-W- L 5 L_ \ A c -(0,2). _ i v 5 u in -10 Aw = -7 A -H A ■ V-(3,-5). 1 A x = 3 L_ V J — 10- A Figure 3.64: Hand-drawn graph of y = — Zx + 2. Check: To graph y = — Zx + 2, enter -7/3*X+2 in Yl in the Y= menu. Select 6:ZStandard from the ZOOM menu, followed by 5:ZSquare from the ZOOM menu to produce the graph shown in Figure 3.66. Answer: llO y y = 2 x ~ 3 -10: (o. m I 2,2 10 = --x + 2 y ^ 1U L 5 L_ \ A r -(0,2). i h \" u in -10 Aw = -7 A -n A ■ V-(3,-5). *\ A x = 3 L_ V J — 10- A Figure 3.65: Hand-drawn graph of y ■lx + 2. Figure 3.66: Select 6:ZStandard from the ZOOM menu, followed by 5:ZSquare from the ZOOM menu to produce the graph of the equa- tion y = —ix + 2. This provides a good match of the hand-drawn graph in Figure 3.65 and our graphing calculator result in Figure 3.66. □ 3.4. SLOPE-INTERCEPT FORM OF A LINE 205 You Try It! EXAMPLE 3. Use the graph of the line in the following figure to find the Use the graph of the line in equation of the line. the figure below to find the equation of the line. V il H / / -If / 10 / / ' / / / / / / / ' If )■ ■ I % v.. N ^ r % N v N S ■s ^^ T > *. T -10 N 1Q U-— io- Solution: Note that the y-intercept of the line is (0,-1) (see Figure 3.67). Next, we try to locate a point on the line that passes directly through a lattice point, a point where a vertical and horizontal grid line intersect. In Figure 3.67, we chose the point (5, 6). Now, starting at the ^-intercept (0, 1), we move up 7 units, then to the right 5 units. Hence, the slope is m = Ay/Aa;, or m = 7/5. y We could also subtract the coordinates of point (0,-1) from the coordinates of point (5, 6) to determine the slope. (-1) 5-0 10 J ■ / / Air c 1 / / r , / ( 5,0! / L \y = V / / /_ / If t - 1( ) -(0,-j 10n 1 — — Figure 3.67: The line has y-intercept (0, —1) and slope 7/5. 206 CHAPTER 3. INTRODUCTION TO GRAPHING Next, state the slope-intercept form, the substitute 7/5 for m and —1 for b. y = mx + b V=-x + {-\) 5 Thus, the equation of the line is y Slope-intercept form. Substitute: 7/5 for to, — 1 for = lx-l. Check: This is an excellent situation to run a check on your graphing calcu- lator. Ploti Plots Plots ■^Y£ = \Yh = ■^Ye = --W? = Figure 3.68: Enter y the Y= menu. 1 in Figure 3.69: Select 6:ZStandard followed by 5:ZSquare (both from the ZOOM menu) to produce this graph. Answer: y = x y 5 When we compare the result in Figure 3.69 with the original hand-drawn graph (see Figure 3.67), we're confident we have a good match. □ You Try It! A swimmer is 50 feet from the beach, and then begins swimming away from the beach at a constant rate of 1.5 feet per second (1.5 ft/s) Express the distance d between the swimmer and the beach in terms of the time t. Applications Let's look at a linear application. EXAMPLE 4. Jason spots his brother Tim talking with friends at the library, located 300 feet away. He begins walking towards his brother at a constant rate of 2 feet per second (2 ft/s). a) Express the distance d between Jason and his brother Tim in terms of the time t. b) At what time after Jason begins walking towards Tim are the brothers 200 feet apart? Solution: Because the distance between Jason and his brother is decreasing at a constant rate, the graph of the distance versus time is a line. Let's begin 3.4. SLOPE-INTERCEPT FORM OF A LINE 207 by making a rough sketch of the line. In Figure 3.70, note that we've labeled what are normally the x- and y-axes with the time t and distance d, and we've included the units with our labels. d(ft) . . (0,300) 200 100 n t(8) 50 100 150 200 Figure 3.70: A plot of the distance d separating the brothers versus time t. Let t = seconds be the time that Jason begins walking towards his brother Tim. At time t = 0, the initial distance between the brothers is 300 feet. This puts the <i-intercept (normally the y-intercept) at the point (0, 300) (see Figure 3.70). Because Jason is walking toward his brother, the distance between the brothers decreases at a constant rate of 2 feet per second. This means the line must slant downhill, making the slope negative, so m = — 2 ft/s. We can construct an accurate plot of distance versus time by starting at the point (0, 300), then descending Ad = —300, then moving to the right At = 150. This makes the slope Ad/ At = -300/150 = -2 (see Figure 3.70). Note that the slope is the rate at which the distance d between the brothers is changing with respect to time t. Finally, the equation of the line is y = mx + b, where m is the slope of the line and b is the y-coordinate (in this case the (i-coordinate) of the point where the graph crosses the vertical axis. Thus, substitute —2 for m, and 300 for b in the slope-intercept form of the line. y = mx + b y = -2x + 300 Slope-intercept form. Substitute: —2 for m, 300 for b. One problem remains. The equation y = —2x + 300 gives us y in terms of x. The question required that we express the distance d in terms of the time t. So, to finish the solution, replace y with d and x with t (check the axes labels 208 CHAPTER 3. INTRODUCTION TO GRAPHING in Figure 3.70) to obtain a solution for part (a): d = -2i + 300 Now that our equation expresses the distance between the brothers in terms of time, let's answer part (b), "At what time after Jason begins walking towards Tim are the brothers 200 feet apart?" To find this time, substitute 200 for d in the equation d = —2t + 300, then solve for t. d = —2t + 300 Distance equation. 200 = -2t + 300 Substitute 200 for d. Solve this last equation for the time t. 200 - 300 = -2t + 300 - 300 Subtract 300 from both sides. — 100 = — It Simplify both sides. -100 -2t Divide both sides by —2. -2 -2 50 = t Simplify both sides. Thus, it takes Jason 50 seconds to close the distance between the brothers to Answer: d = 1.5* + 50 200 feet. □ 3.4. SLOPE-INTERCEPT FORM OF A LINE 209 ti. ;». ;». Exercises ■*j ■*-: ■*; In Exercises 1-6, setup a coordinate system on graph paper, then sketch the line having the given equation. Label the line with its equation. 1. y = |x - 6 2. j/=fz-l 3.y = -^x + 4 4- y = jx 5. y = -^x + 4 6. y = ~lx + 7 In Exercises 7-12, sketch the line with given y-intercept slope. Label the line with the slope-intercept form of its equation. 7. (0,-7), 9/5 8. (0,7), -4/5 9. (0,-1), 6/7 10. (0,1), -7/5 11. (0,-6), 9/7 12. (0,-5), 7/5 In Exercises 13-20, determine the equation of the given line in slope-intercept form. 13. 14. ti 6 \ 6 — -'6-1—1 — t- 210 CHAPTER 3. INTRODUCTION TO GRAPHING 15. 18. "Tvl 6 f 6 \ 6 — 6-1— I -^ 6 J fi fi / / fi 16. 19. I (■ ' / fi f / / / s fi fi' fi fi fi 17. 20. 0' fi fi \ s fi 1 / fi f fi 3.4. SLOPE-INTERCEPT FORM OF A LINE 211 21. Assume that the relationship between an object's velocity and its time is linear. At t = seconds, the object's initial velocity is 20 m/s. It then begins to speed up at a constant rate of 5 meters per second per second. a) Set up a coordinate system, placing the time t on the horizontal axis and the ve- locity v on the vertical axis. Label and scale each axis. Include units in your la- bels. b) Use the initial velocity and the rate at which the object's velocity is increasing to draw the line representing the object's ve- locity at time t. Use the slope-intercept form to determine the equation of the line. c) Replace x and y in the equation found in part (b) with t and v, respectively. Use the result to determine the velocity of the object after 14 seconds. 22. Assume that the relationship between an object's velocity and its time is linear. At t = seconds, the object's initial velocity is 80 m/s. It then begins to lose speed at a constant rate of 4 meters per second per second. a) Set up a coordinate system, placing the time t on the horizontal axis and the ve- locity v on the vertical axis. Label and scale each axis. Include units in your la- bels. b) Use the initial velocity and the rate at which the object's velocity is increasing to draw the line representing the object's ve- locity at time t. Use the slope-intercept form to determine the equation of the line. c) Replace x and y in the equation found in part (b) with t and v, respectively. Use the result to determine the velocity of the object after 13 seconds. 23. A water tank initially (at time t = min) contains 100 gallons of water. A pipe is opened and water pours into the tank at a constant rate of 25 gallons per minute. Assume that the relation between the vol- ume V of water in the tank and time t is linear. a) Set up a coordinate system, placing the time t on the horizontal axis and the vol- ume of water V on the vertical axis. Label and scale each axis. Include units in your labels. b) Use the initial volume of water in the tank and the rate at which the volume of water is increasing to draw the line representing the volume V of water in the tank at time t. Use the slope- intercept form to deter- mine the equation of the line. c) Replace x and y in the equation found in part (b) with t and V, respectively. Use the result to predict how much time must pass until the volume of water in the tank reaches 400 gallons. 24. A water tank initially (at time t = min) contains 800 gallons of water. A spigot is opened at the bottom of the tank and water pours out at a constant rate of 40 gallons per minute. Assume that the re- lation between the volume V of water in the tank and time t is linear. a) Set up a coordinate system, placing the time t on the horizontal axis and the vol- ume of water V on the vertical axis. Label and scale each axis. Include units in your labels. b) Use the initial volume of water in the tank and the rate at which the volume of water is decreasing to draw the line representing the volume V of water in the tank at time t. Use the slope-intercept form to deter- mine the equation of the line. c) Replace x and y in the equation found in part (b) with t and V, respectively. Use the result to predict how much time must pass until the water tank is empty. 212 CHAPTER 3. INTRODUCTION TO GRAPHING s*- $*> £»■ Answers •*$ •*& ■** -10 Ay 10 Ax = «6 / / / /, 2/= la: -6 0(5,3) ■ f x P(0,-6) 10 3. ) J M- r_ 3_ T~t/r\ a \ ) "(0,4); 5 \ T \ T '*-' lAy = —11" \ 10 i -4-a 3 zq t 0(4,-7) Jh / ix — 4 - 10- ' jr - 11 x + 4 y = ji- 1 0(7,5) 11. -10 10 Ay = y = fa; -6 0(7,3) ■ t x P(0,-6) 10 \ 10 1 -lAy = -11 -10 P(0, \ 4) Ax = 7- "111 10 7. -10 :Ay ; y 10+ -id Ax = I / I P(0, 0(7, -7) y = -±fx + 4 9 r7 y = 5^-7 / 0(5, J8)s 10 -7) 13. y = --x y 3 15. y = —-x + 1 17. j/= --x 19. y = -x - 4 5 21. c) 90 m/s 23. c) 12 min 3.5. POINT-SLOPE FORM OF A LINE 213 3.5 Point-Slope Form of a Line In the previous section we learned that if we are provided with the slope of a line and its y-intercept, then the equation of the line is y = mx + 6, where m is the slope of the line and b is the y-coordinate of the line's y-intercept. However, suppose that the y-intercept is unknown? Instead, suppose that we are given a point (xo 1 2/o) on the line and we're also told that the slope of the line is m (see Figure 3.71). * x Slope = m Figure 3.71: A line through the point (xo,yo) with slope m. Let (x,y) be an arbitrary point on the line, then use the points (xo,yo) and (x, y) to calculate the slope of the line. Slope Ay Ax y-yo X — Xq The slope formula. Substitute m for the slope. Use (x, y) and (xo,j/o) to calculate the difference in y and the difference in x. Clear the fractions from the equation by multiplying both sides by the common denominator. m(x — Xo) y-yo _x — Xo m(x- x ) =y-yo (x — xo) Multiply both sides by x — xo- Cancel. Thus, the equation of the line is y — yo = m(x — Xq) 214 CHAPTER 3. INTRODUCTION TO GRAPHING You Try It! Draw the line passing through the point (1,-2) that has slope 3/2, then label it with its equation. Answer: - ' / 5 5 ( r ^ \ L J "J K y + 2=-( X -l) The Point -Slope form of . i line. The equation of the line with slope m that passes through the point (xq ,Vo) is: y-yo = m(x — xq) EXAMPLE 1. Draw the line passing through the point (—3,-1) that has slope 3/5, then label it with its equation. Solution: Plot the point (—3,-1), then move 3 units up and 5 units to the right (see Figure 3.72). To find the equation, substitute (—3,-1) for (xo,yo) and 3/5 for m in the point-slope form of the line. y-y = m(x -Xq) V-(-l) = f(*-(-3)) Point-slope form. Substitute: 3/5 for m, —3 for xq, and —1 for j/q- Simplifying, the equation of the line is y + 1 = |(a; + 3). Ay - ' o y + l = = |( !B + 3 ) A: c = = 5 / = 3 / 5 \ (-3.-1) x, ■ Figure 3.72: The line passing through (—3, —1) with slope 3/5. □ At this point, you may be asking the question "When should I use the slop- intercept form and when should I use the point-slope form?" Here is a good tip. Tip. Which form should I use? The form you should select depends upon the information given. 3.5. POINT-SLOPE FORM OF A LINE 215 1. If you are given the y-intercept and the slope, use y = mx + b. 2. If you are given a point and the slope, use y — yo = m(x — xq). You Try It! EXAMPLE 2. Find the equation of the line passing through the points Find the equation of the line P(— 1, 2) and Q(3, —4). passing through the points P(-2,l) andQ(4,-l). Solution: First, plot the points P(— 1, 2) and <5(3, —4) and draw a line through them (see Figure 3.73). 0(3,-4) Figure 3.73: The line passing through P (— 1, 2) and Q(3, —4). Next, let's calculate the slope of the line by subtracting the coordinates of the point P(— 1, 2) from the coordinates of point Q(3, —4). Slope = — — Ax -4-2 ~3-(-l) -6 3 ~~2 Thus, the slope is —3/2. Next, use the point-slope form y — yo = m(x—xo) to determine the equation of the line. It's clear that we should substitute —3/2 for m. But which of the two points should we use? If we use the point P[— 1, 2) for (xq, yo), we get the answer on the left, but if we use the point Q(3, — 4) for (xo,yo), we get the 216 CHAPTER 3. INTRODUCTION TO GRAPHING answer on the right. y -jOs-C- 1 )) y- (-4) = --(*- 3) At first glance, these answers do not look the same, but let's examine them a bit more closely, solving for y to put each in slope-intercept form. Let's start with the equation on the left. Note that the form y = — \x + \, when compared with the general slope-intercept form y = mx + b, indicates that the y-intercept is (0, 1/2). Examine Figure 3.73. Does it appear that the y-intercept is (0, 1/2)? 1 1 Answer: y = x H — y 3 3 y-2 y-2 2/-2 y-2 + 2 y - 5 (* -M)) (x + l) 3 3 2 3 —x 2 2 3 3 2 X ~ 2 3 3 — x 2 2 3 1 --x+ - 2 2 Using m = —3/2 and (xq, j/o) = (— 1, 2). Simplify. Distribute -3/2. Add 2 to both sides. On the left, simplify. On the right, make equivalent fractions with a common denominator. Simplify. Let's put the second equation in slope-intercept form. y-(-4) = -5(*-3) Using m = —3/2 and (xo,yo) = (3, — y + 4 = -§(*-3) Simplify. y + A = 3 9 --ar+ - 2 2 Distribute -3/2. y + 4- 4 = 3 9 — jc H 2 2 4 Subtract 4 from both sides. 2/ = 3 9 — x H 8 On the left, simplify. On the right, 2 2 2 make equivalent fractions with a common denominator. y = 3 1 — x H — 2 2 Simplify. Thus, both equations simplify to the same answer, y = — Ice + i. This means that the equations y — 2 = — |(x — (—1)) and y — (—4) = — I(cc — 3), though they look different, are the same. □ 3.5. POINT-SLOPE FORM OF A LINE 217 Example 2 gives rise to the following tip. Tip. When finding the equation of a line through two points P and Q, you may substitute either point P or Q for (xq, J/o) m the point-slope form y — yo = m(x — xq). The results look different, but they are both equations of the same line. Parallel Lines Recall that slope is a number that measures the steepness of the line. If two lines are parallel (never intersect), they have the same steepness. Parallel lines. If two lines are parallel, they have the same slope. You Try It! EXAMPLE 3. Sketch the line y 2, then sketch the line passing through the point (—1,1) that is parallel to the line y equation of this parallel line. 2. Find the Solution: Note that y = |x — 2 is in slope-intercept form y = mx + b. Hence, its slope is 3/4 and its ^-intercept is (0, —2). Plot the y-intercept (0, —2), move up 3 units, right 4 units, then draw the line (see Figure 3.74). Find the equation of the line which passes through the point (2, —3) and is parallel to the line y -x + l. *■ X > X Figure 3.74: The line y = jx Figure 3.75: Adding a line parallel toy= fa: -2. The second line must be parallel to the first, so it must have the same slope 3/4. Plot the point (— 1, 1), move up 3 units, right 4 units, then draw the line (see the red line in Figure 3.75). 218 CHAPTER 3. INTRODUCTION TO GRAPHING To find the equation of the parallel red line in Figure 3.75, use the point- slope form, substitute 3/4 for to, then (—1, 1) for (%o,yo). That is, substitute — 1 for xq and 1 for y^. y-yo = m(x-x Q ) y-l = !(*-(-!)) y-l=-(x + l) Point-slope form. Substitute: 3/4 for to, — 1 for xo, and 1 for g/o- Simplify. Plotl PlotE Plots WiB3^4*K-2 Wj = \Yh= We = Figure 3.76: Enter equations of parallel lines. WINDOW Xmin= _ 5 Xmax=5 Xscl=l Vmin= _ 5 Vmax=5 Vscl=l Figure 3.77: Adjust the WINDOW parameters as shown. Check: In this example, we were not required to solve for y, so we can save ourselves some checking work by writing the equation y — 1 = — (x + 1) in the form y = -(x + l) + l by adding 1 to both sides of the first equation. Next, enter each equation as shown in Figure 3.76, then change the WINDOW setting as shown in Figure 3.77. Next, press the GRAPH button, the select 5:ZSquare to produce the image in Figure 3.79. .....jf •// Figure 3.79: Press the GRAPH button then select 5:ZSquare to produce this image. Figure 3.78: Hand-drawn parallel lines. Answer: y = —x y 2 Note the close correlation of the calculator lines in Figure 3.79 to the hand- drawn lines in Figure 3.78. This gives us confidence that we've captured the correct answer. □ 3.5. POINT-SLOPE FORM OF A LINE 219 Perpendicular Lines Two lines are perpendicular if they meet and form a right angle (90 degrees) . For example, the lines £i and £2 in Figure 3.80 are perpendicular, but the lines C\ and £2 in Figure 3.81 are not perpendicular. Figure 3.80: The lines £1 and £2 are perpendicular. They meet and form a right angle (90 degrees). Figure 3.81: The lines C\ and £2 are not perpendicular. They do not form a right angle (90 degrees) . Before continuing, we need to establish a relation between the slopes of two perpendicular lines. So, consider the perpendicular lines £1 and £2 in Figure 3.82. Figure 3.82: Perpendicular lines £1 and £2 Things to note: 1. If we were to rotate line £1 ninety degrees counter-clockwise, then C\ would align with the line £2, as would the right triangles revealing their slopes. Ay mi 2. £1 has slope = — F Ax 1 mi. 220 CHAPTER 3. INTRODUCTION TO GRAPHING 3. £2 has slope Ay Ax -mi 1 mi Slopes of perpendicular lines. If C\ and £2 are perpendicular lines and C\ has slope mi, the £2 has slope — 1/mi. That is, the slope of £2 is the negative reciprocal of the slope of C\. Examples: To find the slope of a perpendicular line, invert the slope of the first line and negate. • If the slope of £1 is 2, then the slope of the perpendicular line £2 is — 1/2. • If the slope of £1 is —3/4, then the slope of the perpendicular line £2 is 4/3. You Try It! Find the equation of the line that passes through the point (—3, 1) and is perpendicular to the line y 1. EXAMPLE 4. Sketch the line y = — |cc — 3, then sketch the line through (2, 1) that is perpendicular to the line y = — fa; — 3. Find the equation of this perpendicular line. Solution: Note that y = — fa; — 3 is in slope-intercept form y = mx + b. Hence, its slope is —2/3 and its y-intercept is (0,-3). Plot the y-intercept (0,-3), move right 3 units, down two units, then draw the line (see Figure 3.83). Figure 3.83: The line y = — fa; — 3. / 1 \x ; / A y = 3 S (2 1) 5 / / -5- Figure 3.84: Adding a line perpen- dicular to y = — \x — 3. Because the line y = -»a:-3 has slope —2/3, the slope of the line perpendicular to this line will be the negative reciprocal of —2/3, namely 3/2. Thus, to draw the perpendicular line, start at the given point (2, 1), move up 3 units, right 2 units, then draw the line (see Figure 3.84). 3.5. POINT-SLOPE FORM OF A LINE 221 To find the equation of the perpendicular line in Figure 3.84, use the point- slope form, substitute 3/2 for to, then (2, 1) for (xo,yo)- That is, substitute 2 for xq, then 1 for y$. y-yo y-i m(x — Xq) Point-slope form. Substitute: 3/2 for m, 2 for xq, and 1 for y$. Check: In this example, we were not required to solve for y, so we can save ourselves some checking work by writing the equation y 3, 1 = -(x 2 V in the form y -{x-2) + \ by adding 1 to both sides of the first equation. Next, enter each equation as shown in Figure 3.85, then select 6:ZStandard to produce the image in Figure 3.86. Ploti Plots Plots \ViB-2^3*X~3 W2B3/-2*<X-2>+i \Vi = --W? = Figure 3.85: Enter equations of per- pendicular lines. Figure 3.86: 6:ZStandard duces two lines that do not perpendicular. pro- look Note that the lines in Figure 3.86 do not appear to be perpendicular. Have we done something wrong? The answer is no! The fact that the calculator's viewing screen is wider than it is tall is distorting the angle at which the lines meet. To make the calculator result match the result in Figure 3.84, change the window settings as shown in Figure 3.87, then select 5:ZSquare from the ZOOM menu to produce the image in Figure 3.88. Note the close correlation of the calculator lines in Figure 3.88 to the hand-drawn lines in Figure 3.84. This gives us confidence that we've captured the correct answer. Answer: y = 2x + 7 □ 222 CHAPTER 3. INTRODUCTION TO GRAPHING WINDOW Xnun= - 5 Xnax=5 Xscl=l Vnun= - 5 Vroax=5 Vscl=l ■i-Xres=l Figure 3.87: Change the WIN- DOW parameters as shown. Figure 3.88: 5:ZSquare produces two lines that do look perpendicu- lar. You Try It! Find an equation that expresses the Fahrenheit temperature in terms of the Celsius temperature. Use the result to find the Fahrenheit temperature when the Celsius temperature is 25° C. Applications Let's look at a real- world application of lines. EXAMPLE 5. Water freezes at 32° F (Fahrenheit) and at 0° C (Celsius). Water boils at 212° F and at 100° C. If the relationship is linear, find an equa- tion that expresses the Celsius temperature in terms of the Fahrenheit tem- perature. Use the result to find the Celsius temperature when the Fahrenheit temperature is 113° F. Solution: In this example, the Celsius temperature depends on the Fahren- heit temperature. This makes the Celsius temperature the dependent variable and it gets placed on the vertical axis. This Fahrenheit temperature is the independent variable, so it gets placed on the horizontal axis (see Figure 3.89). Next, water freezes at 32° F and 0° C, giving us the point (F, C) = (32, 0). Secondly, water boils at 212° F and 100° C, giving us the point (F, C) = (212,100). Note how we've scaled the axes so that each of these points fit on the coordinate system. Finally, assuming a linear relationship between the Celsius and Fahrenheit temperatures, draw a line through these two points (see Figure 3.89). Calculate the slope of the line. Slope formula. AC m = AF 100- -0 212- 100 -32 m = 180 5 m = 9 Use the points (32,0) and (212, 100). to compute the difference in C and F. Simplify. Reduce. You may either use (32,0) or (212,100) in the point-slope form. The point (32,0) has smaller numbers, so it seems easier to substitute (cco,j/o) = (32,0) 3.5. POINT-SLOPE FORM OF A LINE 223 C 120 100 SO GO 40 20 -20 -40 (32,0) (212,100) 40 60 80 100 120 140 160 180 200 220 240 Figure 3.89: The linear relationship between Celsius and Fahrenheit tempera- ture. and m = 5/9 into the point-slope form y — yo = rn(x — Xq). 1 - 2/o = rn(x - Xq) Point-slope form. y-0=^(x-32) Substitute: 5/9 for m, 32 for xo, and for yo- V=\(?- 32) Simplify. However, our vertical and horizontal axes are labeled C and F (see Figure 3.89) respectively, so we must replace y with C and x with F to obtain an equation expressing the Celsius temperature C in terms of the Fahrenheit temperature F. C=^(F-32) (3.1) Finally, to find the Celsius temperature when the Fahrenheit temperature is 113° F, substitute 113 for F in equation (3.1). C C c c 9 5 9 5 9 45 (F-32) (113-32) (81) Equation (3.1). Substitute: 113 for F. Subtract. Multiply. Therefore, if the Fahrenheit temperature is 113° F, then the Celsius tempera- ture is 45° C. Answer: 77° F □ 224 CHAPTER 3. INTRODUCTION TO GRAPHING f> t* t* Exercises «•* -*$ •** In Exercises 1-6, set up a coordinate system on a sheet of graph paper, then sketch the line through the given point with the given slope. Label the line with its equation i point-slope form. 1. m = -5/7, P(-3,4) 4. m = 5/6, P(-3,-l) 2. m = 3/4, P(-2, -4) 5. m = 5/3, P(-l, -4) 3. m = -4/5, P(-2, 4) 6. m = -3/8, P(-4, 0) In Exercises 7-12, set up a coordinate system on a sheet of graph paper, then sketch the line through the given point with the given slope. Label the line with the point-slope form of its equation. 7. P(-4, 0) and Q(4, 3) 10. P(-3, 4) and Q(2, 0) 8. P(-2,4) and 0(2,-1) 11. P(-3, 1) and 0(4,-4) 9. P(-3, -3) and 0(2, 0) 12. P(-l, 0) and 0(4, 3) In Exercises 13-18, on a sheet of graph paper, sketch the given line. Plot the point P and draw a line through P that is parallel to the first line. Label this second line with its equation in point-slope form. 13.y = -fa; + l, P(-3,2) 14. y = -|x, P(-4,0) 15. y=jk P(-4,0) In Exercises 19-24, on a sheet of graph paper, sketch the line passing through the points P and Q. Plot the point R and draw a line through R that is perpendicular to the line passing through the points P and 0- Label this second line with its equation in point-slope form. 19. P(-2, 0), Q(2, -3), and P(-l, 0) 22. P(-4, -4), Q(-l, 3), and P(-4, 2) 20. P(-l,3), 0(2,-2), and P(-1,0) 23. P(-2,3), 0(1,-1), and #(-3,-4) 21. P(-2,-4), 0(1,4), and P(-4,-l) 24. P(-4,4), 0(1,-4), and #(-4,-3) 16. y = fa;- 1, P(-2,-2) 17. y= jx + i, P(-3,0) 18. y = fa; -4, P(-2,-2) 3.5. POINT-SLOPE FORM OF A LINE 225 25. Assume that the relationship between an object's velocity and its time is linear. At 3 seconds, the object's velocity is 50 ft/s. At 14 seconds, the object's velocity is 30 ft/s. a) Set up a coordinate system, placing the time t on the horizontal axis and the velocity v on the vertical axis. La- bel and scale each axis. Include units in your labels. b) Plot the points determined by the given data and draw a line through them. Use the point-slope form of the line to determine the equation of the line. c) Replace x and y in the equation found in part (b) with t and v, respectively, then solve the resulting equation for v. d) Use the result of part (c) to determine the velocity of the object after 6 sec- onds. 26. Water freezes at approximately 32° F and 273° K, where F represents the tempera- ture measured on the Fahrenheit scale and K represents the temperature measured on the Kelvin scale. Water boils at ap- proximately 212° F and 373° K. Assume that the relation between the Fahrenheit and Kelvin temperatures is linear. a) Set up a coordinate system, placing the Kelvin temperature K on the hor- izontal axis and the Fahrenheit tem- perature F on the vertical axis. Label and scale each axis. Include units in your labels. b) Plot the points determined by the given data and draw a line through them. Use the point-slope form of the line to determine the equation of the line. c) Replace x and y in the equation found in part (b) with K and F, respec- tively, then solve the resulting equa- tion for F. d) Use the result of part (c) to deter- mine the Fahrenheit temperature of the object if the Kelvin temperature is 212° K. ;:»■;■*■;*. Answers ■*s ■*■> •*; Ay = 0(4,-1) 6 •4 = -f(a: + 3) 0(3,0) y x y-4 Ux + 2) 226 CHAPTER 3. INTRODUCTION TO GRAPHING 5. 11. y / .y / / \x = rj 7 nfo iv (j T ' \^j 5 Ay- ' z / t / Pi i a\ y ^ ; W + 4=|(x + l) -P(-4,0)- y=§(x + 4) Q(4,3) > a; Alternate answer: y — 3 = | (x — 4) •f(z + 3) Alternate answer: y + 4 = — = (x — 4) 13. ,, = -§35 + 1 \ — ; •■ V P( ^ 9^ *A 0,ZJ 6 (j — i V- 2 = -f(ai + 3) 9. — B J ^ / r )(2,0)6 /• l - -F(-3,-3 i ■i / i — fi y + 3= §(x + 3) 15. Alternate answer: y — |(x — 2) i / ^ / / / / s & > y -A ,c 1 (j -1 y = ix J ' — fi »=!(* + 4) 3.5. POINT-SLOPE FORM OF A LINE 227 17. v = f(a; + 3) 21. 6 •■ 6 V F >{ q rA (3 7-r v u, u; 2/ = J -If) J 2 / 7 r 7 I i ---fl(-4,- A ». =,k=; --. i -h ^ /-- f •--. = _ 7 "^-^ / 6- ' y+l = -|(x + 4) 23. 19. V= f(*+l) 6 ( 6 - f f(- -1 .( ) 6 s s — fi \ \ \ — 6 < \ 6 (j \ pi ' q ' /i\ \ fiJ 5 \ 25. 44.5 ft/s y + 4=f(x + 3) 228 CHAPTER 3. INTRODUCTION TO GRAPHING 3.6 Standard Form of a Line In this section we will investigate the standard form of a line. Let's begin with a simple example. You Try It! Solve x — 2y = 6 for y and plot the result. EXAMPLE 1. Solve the equation 2x + 3y = 6 for y and plot the result. Solution: First we solve the equation 2x + 2>y = 6 for y. Begin by isolating all terms containing y on one side of the equation, moving or keeping all the remaining terms on the other side of the equation. 2x + 3y = 6 2x + 3y - 2x = 6 - 2x 3y = 6 — 2x 3y 6 — 2x T ~ 3 Original equation. Subtract 2x from both sides. Simplify. Divide both sides by 3. Just as multiplication is dis- tributive with respect to ad- dition a{b + c) = ab + ac, so too is division distributive with respect to addition. a + b a b c c c When dividing a sum or a difference by a number, we use the distributive property and divide both terms by that number. On the left, simplify. On the right, divide both terms by 3. G 2x 3 3 9 - 2x 3 Simplify. Finally, use the commutative property to switch the order of the terms on the right-hand side of the last result. Answer: 2x y 2 -— x 3 Add the opposite. Use the commutative property to switch the order. Because the equation 2x + 3j/ = 6 is equivalent to the equation y = — -^x + 2, the graph of 2x + 3y = 6 is a line, having slope m = —2/3 and y-intercept (0, 2). Therefore, to draw the graph of 2x + 3y = 6, plot the y-intercept (0, 2), move down 2 and 3 to the right, then draw the line (see Figure 3.90). □ The form Ax + By = C, where either A = or B = 0, will be handled at the end of this section. In general, unless B = 0, we can always solve the equation Ax + By = C 3.6. STANDARD FORM OF A LINE 229 *■ x Figure 3.90: The graph of 2x + 3y = 6, or equivalently, y = — ^x + 2. for y: Ax + By = c Ax + By - Ax = C- Ax By = C- Ax By C - Ax B B y = C Ax B ~ ~B y = A C ~B X+ B Original equation. Subtract Ax from both sides. Simplify. Divide both sides by B, possible if B ^ 0. On the left, simplify. On the right distribute the B. Commutative property. Note that the last result is in slope-intercept form y = mx + b, whose graph is a line. We have established the following result. Fact. The graph of the equation Ax + By = C, is a line. Important points. A couple of important comments are in order. 1. The form Ax + By = C requires that the coefficients A, B, and C are integers. So, for example, we would clear the fractions from the form 1 2 1 —x -\ — y = — 2 3 y 4 230 CHAPTER 3. INTRODUCTION TO GRAPHING by multiplying both sides by the least common denominator. i2 G* + I y ) = G) 12 Qx + 8y = 3 Note that the coefficients are now integers. 2. The form Ax + By = C also requires that the first coefficient A is non- negative; i.e., A > 0. Thus, if we have — 5a; + 2y = 6, then we would multiply both sides by —1, arriving at: -l(-5x + 2y) = (6)(-l) 5a; — 2y = —6 Note that A = 5 is now greater than or equal to zero. 3. If A, B, and C have a common divisor greater than 1, it is recommended that we divide both sides by the common divisor, thus "reducing" the coefficients. For example, if we have 3x + 12y = -24, then dividing both side by 3 "reduces" the size of the coefficients. 3a; + 12y _ -24 3 ~ "ST x + 4y = -8 Standard form. The form Ax + By = C, where A, B, and C are integers, and A > 0, is called the standard form of a line. Slope-Intercept to Standard Form We've already transformed a couple of equations in standard form into slope- intercept form. Let's reverse the process and place an equation in slope- intercept form into standard form. STANDARD FORM OF A LINE 231 You Try It! EXAMPLE 2. Given the graph of the line in Figure 3.91, find the equation Given the graph of the line of the line in standard form. below, find the equation of the line in standard form. Figure 3.91: Determine the equa- tion of the line. Figure 3.92: The line has y- intercept (0,-3) and slope —5/2. * x Solution: The line intercepts the j/-axis at (0,-3). From (0,-3), move up 5 units, then left 2 units. Thus, the line has slope Ay/Aa: = —5/2 (see Figure 3.92). Substitute —5/2 for m and —3 for b in the slope-intercept form of the line. y = rax + i 5 Slope-intercept form. Substitute: —5/2 for m, —3 for To put this result in standard form Ax + By = C, first clear the fractions by multiplying both sides by the common denominator. 2y 2y 2y —x — 3 2 —x 2 -5a; — 6 Multiply both sides by 2. 2 [3] Distribute the 2. Multiply. That clears the fractions. To put this last result in the form Ax + By = C, we need to move the term —5x to the other side of the equation. 5x + 2y 5x + 2y -5a; — 6 + 5a; -6 Add 5a; to both sides. Simplify. 232 CHAPTER 3. INTRODUCTION TO GRAPHING Answer: 3a; Thus, the standard form of the line is 5x + 2y = —6. Note that all the coeffi- cients are integers and the terms are arranged in the order Ax + By = C, with A>0. □ You Try It! Find the standard form of the equation of the line that passes through the points (-2,4) and (3,-3). Point-Slope to Standard Form Let's do an example where we have to put the point-slope form of a line in standard form. EXAMPLE 3. Sketch the line passing through the points (—3,-4) and (1, 2), then find the equation of the line in standard form. Solution: Plot the points (—3, —4) and (1, 2), then draw a line through them (see Figure 3.93). (-3,-4j Figure 3.93: The line through (—3, —4) and (1,2). Use the points (—3,-4) and (1,2) to calculate the slope. Slope formula. Slope = — — Ax 2 ~(-4) l-(-3) 6 4 3 2 Subtract coordinates of (—3, —4) from the coordinates of (1, 2). Simplify. Reduce. STANDARD FORM OF A LINE 233 Let's substitute (xo,yo) = (1,2) and m = 3/2 in the point-slope form of the line. (Note: Substituting (xo,yo) = (—3,-4) and m = 3/2 would yield the same answer.) y-y = m (x - xq) 1/-2 =|(a;-l) Point-slope form. Substitute: 3/2 for m, 1 for xo, and 2 for j/q. The question requests that our final answer be presented in standard form. First we clear the fractions. 3 3 y — 2 = —x y 2 2 2 [y - 2] = 2 2y - 2[2] = 2 2y - 4 = 3a; - 3 rs 3" — X 2 ~ 2 "3 — X 2 -2 '3" 2 Distribute the 3/2. Multiply both sides by 2. Distribute the 2. Multiply. Now that we've cleared the fractions, we must order the terms in the form Ax + By = C . We need to move the term 3a; to the other side of the equation. 1y — 4 — Sx = Sx — 3 — Sx Subtract 3a; from both sides. — Sx + 2y — 4 = —3 Simplify, changing the order on the left-hand side. To put this in the form Ax + By = C, we need to move the term —4 to the other side of the equation. -Sx + 2y-A + A = -3 + 4 -3x + 2y = 1 Add 4 to both sides. Simplify. It appears that —3a; + 2y = 1 is in the form Ax + By = C. However, standard form requires that A > 0. We have A = —3. To fix this, we multiply both sides by —1. If we fail to reduce the slope to lowest terms, then the equation of the line would be: y-2 = ^(x-l) Multiplying both sides by 4 would give us the result Ay — 8 = 6a; — 6, or equivalently: -6a; + Ay = 2 This doesn't look like the same answer, but if we divide both sides by —2, we do get the same result. 3a; — 2y -1 This shows the importance of requiring A > and "re- ducing" the coefficients A, B, and C. It allows us to com- pare our answer with our col- leagues or the answers pre- sented in this textbook. -1 [-3s + 2y] = -1 [1] 3x- 2y = -1 Multiply both sides by — 1. Distribute the — 1. Thus, the equation of the line in standard form is 3a; — 2y = — 1. Answer: 7x + 5y = 6 □ 234 CHAPTER 3. INTRODUCTION TO GRAPHING Intercepts We've studied the y-intercept, the point where the graph crosses the y-axis, but equally important are the ^-intercepts, the points where the graph crosses the a;-axis. In Figure 3.94, the graph crosses the x-axis three times. Each of these crossing points is called an rc-intercept. Note that each of these x-intercepts has a y-coordinate equal to zero. This leads to the following rule. x-Intercepts. To find the x-intercepts of the graph of an equation, substitute y = into the equation and solve for x. Similarly, the graph in Figure 3.95 crosses the y-axis three times. Each of these crossing points is called a y-intercept. Note that each of these y-intercepts has an ^-coordinate equal to zero. This leads to the following rule. y-Intercepts. To find the y-intercepts of the graph of an equation, substitute x = into the equation and solve for y. Figure 3.94: Each x-intercept has a y-coordinate equal to Figure 3.95: Each y-intercept has an cc-coordinate equal to zero. Let's put these rules for finding intercepts to work. You Try It! Find the x- and y-intercepts of the line having equation 3x + 4y = -12. Plot the intercepts and draw the line. EXAMPLE 4. Find the x- and y-intercepts of the line having equation 2x — 3y = 6. Plot the intercepts and draw the line. STANDARD FORM OF A LINE 235 Solution: We know that the graph of 2x — 3y = 6 is a line. Furthermore, two points completely determine a line. This means that we need only plot the x- and y-intercepts, then draw a line through them. To find the ^-intercept of 2x — 3y 6, substitute for y and solve for x. To find the y-intercept of 2x — 3y = 6, substitute for x and solve for y. 2x — 3y 2a; -3(0) 2x 2x ~2 x 6 6 6 6 2 3 2x - -3y = 6 2(0)- -3p = 6 -3y = 6 -3y 6 -3 -3 y = = -2 Thus, the ^-intercept of the line is (3,0). Thus, the y-intercept of the line is (0,-2). Plot the x-intercept (3, 0) and the y-intercept (0, —2) and draw a line through them (see Figure 3.96). 2x — 3y = 6 * x Figure 3.96: The graph of 2x — 3y = 6 has intercepts (3, 0) and (0, —2) Answer: x-intercept: (—4, 0) y-intercept: (0, —3) V 5 ( -4 M 11 ') -5- 5 fn N (0, ^ , >) 5 3x + 4y = -12 □ EXAMPLE 5. Sketch the line 4x + 3y = 12, then sketch the line through the point (—2, —2) that is perpendicular to the line 4x + 3y = 12. Find the equation of this perpendicular line. Solution: Let's first find the x- and y-intercepts of the line 4x + 3y = 12. You Try It! Find the equation of the line that passes through the point (3, 2) and is perpendicular to the line 6x — by = 15. 236 CHAPTER 3. INTRODUCTION TO GRAPHING To find the x-intercept of the line Ax + 3y = 12, substitute for y and solve for x. Ax + 3y = 12 4a; + 3(0) = 12 Ax = 12 Ax _ 12 T = T x = 3 Thus, the x-intercept of the line is (3,0). To find the y-intercept of the line Ax + 3y = 12, substitute for x and solve for y. Ax + 3y = 12 4(0) + 3y = 12 3j/ = 12 3y = 12 3 3 y = A Thus, the j/-intercept of the line is (0,4). You could also solve for y to Plot the intercepts and draw a line through them. Note that it is clear from put 3a; + Ay = 12 in slope- the graph that the slope of the line 3a; + Ay = 12 is —4/3 (see Figure 3.97). intercept form in order to de- termine the slope. y y Figure 3.97: The graph of Ax+3y = 12 has intercepts (3,0) and (0,4) and slope —4/3. * x Figure 3.98: The slope of the per- pendicular line is the negative re- ciprocal of —4/3, namely 3/4. Because the slope of 3x + Ay = 12 is —4/3, the slope of a line perpendicular to 3.t + Ay = 12 will be the negative reciprocal of —4/3, namely 3/4. Our perpendicular line has to pass through the point (—2, —2). Start at (—2, —2), move 3 units upward, then 4 units to the right, then draw the line. It should appear to be perpendicular to the line 3x + Ay = 12 (see Figure 3.98). Finally, use the point-slope form, m = 3/4, and (xo,yo) = (—2,-2) to STANDARD FORM OF A LINE 237 determine the equation of the perpendicular line. y-y Q = m(x - x ) Point-slope form. y - (-2) = -{x- (-2)) Substitute: 3/4 for to, -2 for x , y + 2 = -(x + 2) and —2 for j/o- Simplify. Let's place our answer in standard form. Clear the fractions. Distribute 3/4. 3 6 y + 2 = -x+- 4 [y + 2] = 4 r3 6i -x+ - [4 4j [3 1 [fil T x + 4 [A [ij Multiply both sides by 4. Ay + 4 [2] = 4 -a; + 4 - Distribute the 4. 4y + 8 = 3.T + 6 Multiply. Rearrange the terms to put them in the order Ax + By = C. Ay + 8 - 3a; = 3a; + 6 - 3x -3x + Ay + 8 = 6 -3x + Ay + 8-8 = 6-8 -3x + Ay = -2 -l(-3x + Ay) = -l(-2) 3x - Ay = 2 Subtract 3a; from both sides. Simplify. Rearrange on the left. Subtract 8 from both sides. Simplify. Multiply both sides by —1. Distribute the — 1. Hence, the equation of the perpendicular line is 3a; — Ay = 2. Answer: 5x + 6y = 27 □ Horizontal and Vertical Lines Here we keep an earlier promise to address what happens to the standard form Ax + By = C when either A = or B = 0. For example, the form 3a; = 6, when compared with the standard form Ax + By = C, has B = 0. Similarly, the form 2y = —12, when compared with the standard form Ax + By = C, has A = 0. Of course, 3a; = 6 can be simplified to x = 2 and 2y = — 12 can be simplified to y = —6. Thus, if either A = or B = 0, the standard form Ax + By = C takes the form x = a and y = b, respectively. As we will see in the next example, the form x = a produces a vertical line, while the form y = b produces a horizontal line. 238 CHAPTER 3. INTRODUCTION TO GRAPHING You Try It! Sketch the graphs of x and y = 2. EXAMPLE 6. Sketch the graphs of x = 3 and y = -3. Solution: To sketch the graph of x = 3, recall that the graph of an equation is the set of all points that satisfy the equation. Hence, to draw the graph of x = 3, we must plot all of the points that satisfy the equation x = 3; that is, we must plot all of the points that have an x-coordinate equal to 3. The result is shown in Figure 3.99. Secondly, to sketch the graph of y = —3, we plot all points having a y- coordinate equal to —3. The result is shown in Figure 3.100. y y ->■ X *■ X Figure 3.99: The graph of x a vertical line. 3 is Figure 3.100: The graph of y is a horizontal line. Things to note: A couple of comments are in order regarding the lines in Figures 3.99 and 3.100. 1. The graph of x = 3 in Figure 3.99, being a vertical line, has undefined slope. Therefore, we cannot use either of the formulae y = mx + b or y — j/o = m(x — xq) to obtain the equation of the line. The only way we can obtain the equation is to note that the line is the set of all points (x, y) whose ^-coordinate equals 3. 2. However, the graph of y = —3, being a horizontal line, has slope zero, so we can use the slope-intercept form to find the equation of the line. Note that the y-intercept of this graph is (0,-3). If we substitute these numbers into y = mx + b, we get: y = mx + b y = 0x + (-3) y = -3 Slope-intercept form. Substitute: for m, —3 for b. Simplify. Answer: 3.6. STANDARD FORM OF A LINE 239 However, it is far easier to just look at the line in Figures 3.100 and note that it is the collection of all points (x,y) with y = 3. x = -2 t>-' y = 2 -*■ x □ 240 CHAPTER 3. INTRODUCTION TO GRAPHING f> t* t* Exercises ■** -** ■** In Exercises 1-6, place the given standard form into slope-intercept form and sketch its graph. Label the graph with its slope-intercept form. 1.4x-3y = 9 2.2x-3y = 3 3. 3x-2y = 6 4. 5x - 3y = 3 5. 2x + 3y = 12 6. 3x + Ay = 8 In Exercises 7-10, determine an equation of the given line in standard form. 7. 9. 8. y 6- fi f fi 2 r' 1 / \ fi S f \ \ \ s \ 6- \ 6- fi ^fi "" fi 10. -6- t / ~i ' ~7_ j -6 i 6 / frv. 3.6. STANDARD FORM OF A LINE 241 In Exercises 11-16, sketch the line passing through the points P and Q. Label the line with its equation in standard form. 11. P(-l,4) andQ(2,-4) 12. P(-l,4) and (5(3,1) 13. P(-l,-l) and Q(3,4) 14. P(2,-l) andQ(4,4) 15. P(-3,l) and Q(2, -3) 16. P(-4,3) and Q(0,0) In Exercises 17-22, plot the x- and y-intercepts of the line having the given equation, then draw the line through the intercepts and label it with its equation. 17. 2x - 5y = 10 18. 2x + 3y = -6 19. 3x-2y = 6 20. 3x - Ay = 12 21. 2x + 3y = 6 22. 2x-3y = -6 23. Find an equation of the line (in stan- dard form) that passes through the point P(— 1,5) that is parallel to the line Ax + 5y = -20. 24. Find an equation of the line (in stan- dard form) that passes through the point P(— 3,2) that is parallel to the line 3x + 5j/ = -15. 25. Find an equation of the line (in stan- dard form) that passes through the point P(— 1,— 2) that is perpendicular to the line 5x + 2y = 10. 26. Find an equation of the line (in stan- dard form) that passes through the point P(— 1,— 2) that is perpendicular to the line 2x + 5y = -10. 27. Find an equation of the line (in stan- dard form) that passes through the point P(— 4, — 5) that is perpendicular to the line Ax + 3y = —12. 28. Find an equation of the line (in stan- dard form) that passes through the point P(— 2,— 3) that is perpendicular to the line 3x + 2y = 6. 29. Find an equation of the line (in stan- dard form) that passes through the point P(— 3,2) that is parallel to the line 5x + Ay = 20. 30. Find an equation of the line (in stan- dard form) that passes through the point P(— 1,3) that is parallel to the line 3x + 5y = -15. 31 32 Sketch the equation of the horizontal line passing through the point P(— 5, — 4) and label it with its equation. Sketch the equation of the vertical line passing through the point P(— 4, 4) and label it with its equation. 33 34. Sketch the equation of the vertical line passing through the point P(— 2, —4) and label it with its equation. Sketch the equation of the vertical line passing through the point P(l,3) and la- bel it with its equation. 242 CHAPTER 3. INTRODUCTION TO GRAPHING £»• s*> £*■ Answers >•* >*$ •** 11. I / / \a y i *■ H: 6 A / = I i / / ( ). / ' \ / /_ fi y = ~x - 3 ¥i H-U- -^ t I \ %. 6 A 6 L_ _| V \ --0(2,- 6- U)X iJLl ix + 3y = 4 — 6 / / / / / z\ a; ^ 2' / (j "(2,0) fi a j/ - ; -(0,-3)- fi »=|*-3 6y = -fa; + 4 7. 4a; + 5y = 9. 2a;-52/= 15 13. — B-f-r-r- .(15 zH VWi ^J 6 (i f v i -A r v A i A ; — fi 15. 5x — Ay = — 1 — 6 P( "i 1 "l r-^ a, i; 6 (i f uo q~\ 1 — fi 4a; + 5y = -7 3.6. STANDARD FORM OF A LINE 243 17. (0,-2)- 2x - by = 10 * x (5,0) 23. Ax + 5y = 21 tP(- i, b) fi fi Ax + 5y = -20 19. 25. G ! (j i" >.( ») -6 i- C( i - q ) ^ j, u 6 3cc — 2y = 6 2x — 5y R t G \ \ \ \ \ \ \ --- -- fi v .--' fi \ F f-l.'-Vl \ \ fi \ hx + 2y = 10 21. 27. -64 (0,2)- (3,0)- 2a; + 3j/ = 6 244 CHAPTER 3. INTRODUCTION TO GRAPHING 29. 33. 5x + 4y = -7 % = 20 y 6 f> i 9 A \ z > ^y fi 31. (-5,-4> -■y = —4 JLJ Chapter 4 Systems of Linear Equations In 1801, Carl Frederick Gauss (1777-1885) computed the orbit of the newly discovered planetoid Ceres from just a few observations by solving a system of equations. He invented a method (called Gaussian elimination) that is still used today. Solving systems of equations has been a subject of study in many other cultures. The ancient Chinese text Jiuzhang Suanshu (translated as Nine Chapters of Mathematical Art) written during the Han dynasty (206 BC-220 AD) describes 246 problems related to practical situations such as land measurement, construction, and commerce. Here is one of the problems described in the text: "One pint of good wine costs 50 gold pieces, while one pint of poor wine costs 10. Two pints of wine are bought for 30 gold pieces. How much of each kind of wine was bought?" This problem can be solved by using a system of equations. In this chapter, we will learn how to solve systems of linear equations by using a variety of methods, including Gaussian elimination. 245 246 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 8- r — 3?/ = - y (3,4) 2x + 32/ = 18 -9, > < L_ 2 4.1 Solving Systems by Graphing In this section we introduce a graphical technique for solving systems of two linear equations in two unknowns. As we saw in the previous chapter, if a point satisfies an equation, then that point lies on the graph of the equation. If we are looking for a point that satisfies two equations, then we are looking for a point that lies on the graphs of both equations; that is, we are looking for a point of intersection. For example, consider the the two equations Figure 4.1: The point of in- tersection is a solution of the system of linear equations. x 2x 32/ 32/ -9 18. which is called a system of linear equations. The equations are linear equations because their graphs are lines, as shown in Figure 4.1. Note that the two lines in Figure 4.1 intersect at the point (3,4). Therefore, the point (3,4) should satisfy both equations. Let's check. Substitute 3 for x and 4 for y. Substitute 3 for x and 4 for y. x - 3y = -9 3 - 3(4) = -9 3-12 = -9 -9 = -9 2a; + 32/ 2(3) + 3(4) 6+12 18 18 18 18 18 Hence, the point (3, 4) satisfies both equations and is called a solution of the system. Solution of a linear system. A point (x, y) is called a solution of a system of two linear equations if and only if it satisfied both equations. Furthermore, because a point satisfies an equation if and only if it lies on the graph of the equation, to solve a system of linear equations graphically, we need to determine the point of intersection of the two lines having the given equations. You Try It! Solve the following system of equations: 2x by V -10 x- 1 Let's try an example. EXAMPLE 1. Solve the following system of equations: 3x + 2y = 12 y = x + 1 (4.1) Solution: We are looking for the point (x, y) that satisfies both equations; that is, we are looking for the point that lies on the graph of both equations. Therefore, the logical approach is to plot the graphs of both lines, then identify the point of intersection. First, let's determine the x- and 2/-intercepts of 3a: + 2y = 12. 4.1. SOLVING SYSTEMS BY GRAPHING 247 To find the ir-intercept, let y = 0. To find the y- intercept, let x = 0. 3x + 2y 3x + 2(0) : 3x 12 12 12 3x + 2y = 12 3(0) + 2y = 12 2y=12 y = 6 Hence, the ^-intercept is (4,0) and the y-intercept is (0,6). These intercepts are plotted in Figure 4.2 and the line 3x + 2y = 12 is drawn through them. Comparing the second equation y = x + 1 with the slope-intercept form y = mx + b, we see that the slope is m = 1 and the y-intercept is (0, 1). Plot the intercept (0, 1), then go up 1 unit and right 1 unit, then draw the line (see Figure 4.3). 3x + 2y = 12 Figure 4.2: Drawing the graph of 3x + 2y= 12. Figure 4.3: Drawing the graph of y = x + 1. We are trying to find the point that lies on both lines, so we plot both lines on the same coordinate system, labeling each with its equation (see Figure 4.4). It appears that the lines intersect at the point (2, 3), making (x, y) = (2, 3) the solution of System 4.1 (see Figure 4.4). Check: To show that (x 1 y) = (2, 3) is a solution of System 4.1, we must show that we get true statements when we substitute 2 for x and 3 for y in both equations of System 4.1. 248 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS Answer: (5,4) * x 3x + 2y = 12 Figure 4.4: The coordinates of the point of intersection is the solution of System 4.1. Substituting 2 for x and 3 for y in 3x + 2y = 12, we get: 3x + 2y = 12 3(2) + 2(3) = 12 6 + 6 = 12 12 = 12 Hence, (2, 3) satisfies the equation 3x + 2y = 12. Substituting 2 for x and 3 for y in y = x + 1, we get: y = x + 1 3 = 2 + 1 3 = 3 Hence, (2,3) satisfies the equation y = x + 1. Because (2, 3) satisfies both equations, this makes (2, 3) a solution of System 4.1. □ You Try It! Solve the following system of equations: -4x 3y 2y 12 EXAMPLE 2. Solve the following system of equations: 3a; — by = — 15 2x + y = -4 (4.2) Solution: Once again, we are looking for the point that satisfies both equations of the System 4.2. Thus, we need to find the point that lies on the graphs of both lines represented by the equations of System 4.2. The approach will be to graph both lines, then approximate the coordinates of the point of intersection. First, let's determine the x- and y-intercepts of 3x — 5y = —15. 4.1. SOLVING SYSTEMS BY GRAPHING 249 To find the x-intercept, let y = 0. To find the y-intercept, let x = 0. 3x — by = -15 32-5(0) = -15 3x = -15 x = -5 3x - - by = -15 3(0)- -by = -15 -by = -15 V = 3 Hence, the ^-intercept is (—5,0) and the y-intercept is (0,3). These intercepts are plotted in Figure 4.5 and the line 3a; — by = — 15 is drawn through them. Next, let's determine the intercepts of the second equation 2x + y = —4. To find the x-intercept, let y = 0. To find the y-intercept, let x = 0. 2x + y = -4 2a; + = -4 2a; = -4 x = -2 2a; + y = -4 2(0) +y = -4 V = -4 Hence, the a;-intercept is (—2, 0) and the y-intercept is (0, —4). These intercepts are plotted in Figure 4.6 and the line 2x -\- y = —4 is drawn through them. V 3x — by - ^ ^ *7 (0,3) . ^ S ^ .(-5,0) ^ . « 1 11^ 1 I A* X>' '- 1U ^ J — 10- -15 y y^j- r 10 J 3 v 3 v 3 v 3 v V - -1U (— 2,UJ ^ 1U v. _5 (0,-4) - *$ V 3 v J — 10^ 3 Later in this section we will learn how to use the intersect utility on the graphing calculator to obtain a much more accurate approximation of the actual solution. Then, in Sections 4.2 and 4.3, we'll show how to find the exact solution. Figure 4.5: the line 3a; - Drawing the graph of -by = -15. 2a; + y = -4 Figure 4.6: Drawing the graph of the line 2x + y = —4. To find the solution of System 4.2, we need to plot both lines on the same coordinate system and determine the coordinates of the point of intersection. Unlike Example 1, in this case we'll have to be content with an approximation of these coordinates. It appears that the coordinates of the point of intersection are approximately (—2.6, 1.4) (see Figure 4.7). Check: Because we only have an approximation of the solution of the system, we cannot expect the solution to check exactly in each equation. However, we do hope that the solution checks approximately. 250 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS tkt r 1(p " ■1:1; .)(/ - 3 : ^ v ^ 3 v .^ 3 __,' \ *' (-2.6,1.4): r^ : J 1 ^ \ 1J .s \ ~¥L y 3 _ iu ^ v 3_ \ 5 :v _3 _ v J — 10^- 3 2x y Figure 4.7: The approximate coordinates of the point of intersection are (-2.6,1.4). Substitute (x, y) = (—2.6,1.4) into the first equation of System 4.2. 3a; — 5y = -15 3(-2.6) - 5(1.4) = -15 -7.8-7 = -15 -14.8 = -15 Substitute {x,y) = (—2.6,1.4) into the second equation of System 4.2. 2x + y = -4 2(-2.6) + 1.4 = -4 -5.2+1.4 = -4 -3.8 = -4 Approximate answer: (-2.7,-0.4) Note that (x,y) = (-2.6,1.4) does not check exactly, but it is pretty close to being a true statement. Again, note that (a;, y) = (—2.6,1.4) does not check ex- actly, but it is pretty close to being a true statement. Because (x,y) = (—2.6,1.4) very nearly makes both equations a true state- ment, it seems that (x, y) = (—2.6, 1.4) is a reasonable approximation for the solution of System 4.2. □ You Try It! Exceptional Cases Most of the time, given the graphs of two lines, they will intersect in exactly one point. But there are two exceptions to this general scenario. Solve the following system of EXAMPLE 3. Solve the following system of equations: equations: x-y -2x + 2y 2x + 3y 2x + 3y 6 -6 (4.3) 4.1. SOLVING SYSTEMS BY GRAPHING 251 Solution: Let's place each equation in slope-intercept form by solving each equation for y. Solve 2x + 3y = 6 for y: 2x + 3y = 6 2x + 3y-2x = 6-2x 3y = 6 — 2x 3y 6 — 2x ~3 ~ 3 2 y = -o x + Solve 2x + 3y = —6 for y: 2x + 3y = -6 2x + 3y — 2x = -6 -2a; 3y = -6- 2x 3y -6 -2x 3 3 y Comparing y = (— 2/3)a; + 2 with the slope-intercept form y = mx + b tells us that the slope is m = —2/3 and the y-intercept is (0, 2). Plot the intercept (0, 2), then go down 2 units and right 3 units and draw the line (see Figure 4.8). Comparing y = (— 2/3)a: — 2 with the slope-intercept form y = mx + b tells us that the slope is m = —2/3 and the y-intercept is (0,-2). Plot the intercept (0, —2), then go down 2 units and right 3 units and draw the line (see Figure 4.9). 2x + 3y Figure 4.8: Drawing the graph of the line 2x + 3y = 6. Figure 4.9: Drawing the graph of the line 2x + 3y = —6. To find the solution of System 4.3, draw both lines on the same coordinate system (see Figure 4.10). Note how the lines appear to be parallel (they don't intersect). The fact that both lines have the same slope —2/3 confirms our suspicion that the lines are parallel. However, note that the lines have differ- ent jy-intercepts. Hence, we are looking at two parallel but distinct lines (see Figure 4.10) that do not intersect. Hence, System 4.3 has no solution. 252 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 2x + 3y = 6 *■ x 2x + 3y = -6 Figure 4.10: The lines 2x + 3y = 6 and 2x + 3y = —6 are parallel, distinct lines. Answer: No solution. □ You Try It! Solve the following system of EXAMPLE 4. Solve the following system of equations: equations: x-y = 3 -2x + 2y = -6 -6x + 3y = -12 2x - y = 4 (4.4) Figure 4.11: x — y = 3 and — 2x + 2 y = — 6 are the same line. Solution: Let's solve both equations for y. Solve x — y = 3 for y: x-y = 3 X - y — x = 3 — x —y = —x + 3 -l(-y) = -l(-x + 3) Solve -2x + 2y = - 6 for y: -2x + 2y = -6 -2x + 2y + 2x = -6 + 2.x 2y = 2a; -6 2y 2x -6 2 2 V = x — 3 y = x Both lines have slope m = 1, and both have the same y-intercept (0,-3). Hence, the two lines are identical (see Figure 4.11). Hence, System 4.4 has an infinite number of points of intersection. Any point on either line is a solution of the system. Examples of points of intersection (solutions satisfying both equations) are (0,-3), (1,-2), and (3,0). 4.1. SOLVING SYSTEMS BY GRAPHING 253 Alternate solution: A much easier approach is to note that if we divide both sides of the second equation —2x + 2y = —6 by —2, we get: Second equation in System 4.4. Divide both sides by —2. Distribute —2. Simplify. -2x + 2y = -6 -2x + 2y -6 -2 -2 -2x 2y -2 + -2 ~ -6 ^2 x-y Hence, the second equation in System 4.4 is identical to the first. Thus, there are an infinite number of solutions. Any point on either line is a solution. Answer: There are an infinite number of solutions. The lines are identical, so any point on either line is a solution. Examples 1, 2, 3, and 4 lead us to the following conclusion. Number of solutions of a linear system. When dealing with a system of two linear equations in two unknowns, there are only three possibilities: 1. There is exactly one solution. 2. There are no solutions. 3. There are an infinite number of solutions. □ Solving Systems with the Graphing Calculator We've already had experience graphing equations with the graphing calculator. We've also used the TRACE button to estimate points of intersection. How- ever, the graphing calculator has a much more sophisticated tool for finding points of intersection. In the next example we'll use the graphing calculator to find the solution of System 4.1 of Example 1. You Try It! EXAMPLE 5. Use the graphing calculator to solve the following system of Solve the following system of equations: equations: 3x + 2?/=12 (45) 2x-5y = 9 V = x + 1 y = 2x-5 Solution: To enter an equation in the Y= menu, the equation must first be 254 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS solved for y. Hence, we must first solve 3a; + 2y = 12 for y. Having the calculator ask "First curve," "Second curve," when there are only two curves on the screen may seem annoying. However, imagine the situation when there are three or more curves on the screen. Then these questions make good sense. You can change your selection of "First curve" or "Second curve" by using the up-and-down arrow keys to move the cursor to a different curve. 3a; 2y = 12 2y = 12- - 3a; 2y 12 — 3a; 2 2 12 3.r y = T ~ T y Original equation. Subtract 3a; from both sides of the equation. Divide both sides by 2. On the left, simplify. On the right, distribute division by 2. Simplify. We can now substitute both equations of System 4.5 into the Y= menu (see Figure 4.12). Select 6:ZStandard from the ZOOM menu to produce the graphs shown in Figure 4.13. Ploti Plots Plot3 sViB6-3^2*X ■^eBX+1 \Vi=W \Yh = ■^Ye = ^Vfi = sY? = Figure 4.12: Enter System 4.5 equations into the Y= menu. Figure 4.13: Select 6:ZStandard to produce the graphs of the sys- tem (4.5) equations. The question now becomes "How do we calculate the coordinates of the point of intersection?" Look on your calculator case just above the TRACE button on the top row of buttons, where you'll see the word CA1C, painted in the same color as the 2ND key. Press the 2ND key, then the TRACE button, which will open the CALCULATE menu shown in Figure 4.14. Select 5:intersect. The result is shown in Figure 4.15. The calculator has placed the cursor on the curve y = 6 — (3/2)x (see upper left corner of your viewing screen), and in the lower left corner the calculator is asking you if you want to use the selected curve as the "First curve." Answer "yes" by pressing the ENTER button. The calculator responds as shown Figure 4.16. The cursor jumps to the curve y = x + 1 (see upper left corner of your viewing window), and in the lower left corner the calculator is asking you if you want to use the selected curve as the "Second curve." Answer "yes" by pressing the ENTER key again. The calculator responds as shown Figure 4.17, asking you to "Guess." In this case, leave the cursor where it is and press the ENTER key again to signal the calculator that you are making a guess at the current position of the cursor. The result of pressing ENTER to the "Guess" question in Figure 4.17 is shown in Figure 4.18, where the calculator now provides an approximation 4.1. SOLVING SYSTEMS BY GRAPHING 255 MMMslla ualue 2Tzero 3:nininun 4:naxinur v i 5: intersect 7:XfCx>dx Figure 4.14: Press 2ND, then TRACE to open the CALCU- LATE menu. Then select 5:in- tersect to produce the screen in Figure 4.15. V1=G-3/£*KF y^ i i y^ FiKSt CUKV4? H=0 v=6 -■ Figure 4.15: Press the ENTER key on your calculator to say "yes" to the "First curve" selection. VE=H+i Second curve? H=0 IV: Figure 4.16: Press the ENTER key on your calculator to say "yes" to the "Second curve" selection. VE=H+i V Guess? H=0 v=i -■ Figure 4.17: Press the ENTER key to signal the calculator that you are satisfied with the current position of the cursor as your guess. of the the coordinates of the intersection point on the bottom edge of the viewing window. Note that the calculator has placed the cursor on the point of intersection in Figure 4.17 and reports that the approximate coordinates of the point of intersection are (2, 3). Intersection K=2 V=3 -■ Figure 4.18: Read the approximate coordinates of the point of intersection along the bottom edge of the viewing window. In later sections, when we investigate the intersection of two graphs having more than one point of intersection, guessing will become more important. In those future cases, we'll need to use the left-and-right arrow keys to move the cursor near the point of intersection we wish the calculator to find. 256 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS Reporting your solution on your homework. In reporting your solution on your homework paper, follow the Calculator Submission Guidelines from Chapter 3, Section 2. Make an accurate copy of the image shown in your viewing window. Label your axes x and y. At the end of each axis, put the appropriate value of Xmin, Xmax, Ymin, and Ymax reported in your calculator's WINDOW menu. Use a ruler to draw the lines and label each with their equations. Finally, label the point of intersection with its coordinates (see Figure 4.19). Unless instructed otherwise, always report every single digit displayed on your calculator. -10 + 3x + 2y = 12 Figure 4.19: Reporting your result on your homework paper. Approximate answer: (2,-1) □ You Try It! Solve the following system of equations: y 6 Sometimes you will need to adjust the parameters in the WINDOW menu so that the point of intersection is visible in the viewing window. EXAMPLE 6. Use the graphing calculator to find an approximate solution of the following system: y = -~x + 7 3 7 ^ 6 ) y = —x — 5 y 5 Solution: Each equation of System 4.6 is already solved for y, so we can proceed directly and enter them in the Y= menu, as shown in Figure 4.20. Select 6:ZStandard from the ZOOM menu to produce the image shown in Figure 4.21. 4.1. SOLVING SYSTEMS BY GRAPHING 257 Ploti Plots Plots WiB-2/7*X+7 W2B3/-5+X-5 ^Ys = \Vh = ^7 = Figure 4.20: Enter the equations of System 4.6. Figure 4.21: Select 6:ZStandard to produce this window. Obviously, the point of intersection is off the screen to the right, so we'll have to increase the value of Xmax (set Xmax = 20) as shown in Figure 4.22. Once you have made that change to Xmax, press the GRAPH button to produce the image shown in Figure 4.23. WINDOW Xnun=-10 Xnax=2@ Xscl=l Wiin=-10 Wiax=l@ Vscl=l ■i-Xres=l Figure 4.22: Change xmax to 20. Figure 4.23: Press the GRAPH button to produce this window. Now that the point of intersection is visible in the viewing window, press 2ND CALC and select 5:intersect from the CALCULATE menu (see Figure 4.25) Make three consecutive presses of the ENTER button to respond to "First curve," "Second curve," and "Guess." The calculator responds with the im- age in Figure 4.25. Thus, the solution of System 4.6 is approximately (x, y) w (13.54837,3.1290323). Warning! Your calculator is an approximating machine. It is quite likely that your solutions might differ slightly from the solution presented in Figure 4.25 in the last 2-3 places. Reporting your solution on your homework. In reporting your solution on your homework paper, follow the Calculator Submission Guidelines from Chapter 3, Section 2. Make an accurate copy of the image shown in your viewing window. Label your axes x and y. At the end of each axis, put the appropriate value of Xmin, Xmax, Ymin, and Ymax reported in your 258 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS MMMslla ualue 2Tzero 3:nininun 4:naxinur v i 5: intersect 7:XfCx>dx Figure 4.24: Press 2ND CALC to open the CALCULATE menu. Select 5:intersect to find the point of intersection. Intersection K=13.£HB3B7 3.i£903£3 Figure 4.25: Three consecutive presses of the ENTER key produce the coordinates shown at the bot- tom of the viewing window. calculator's WINDOW menu. Use a ruler to draw the lines and label each with their equations. Finally, label the point of intersection with its coordinates (see Figure 4.26). Unless instructed otherwise, always report every single digit displayed on your calculator. (13.548387,3.1290323) Figure 4.26: Reporting your result on your homework paper. Approximate answer: (-4.2,-0.4) □ 4.1. SOLVING SYSTEMS BY GRAPHING 259 i*. ;». ;». Exercises ■*j ■*-: •*; In Exercises 1-6, solve each of the given systems by sketching the lines represented by each equation in the system, then determining the coordinates of the point of intersection. Each of these problems have been designed so that the coordinates of the intersection point are integers. Check your solution. 3a; -Ay-- = 24 y- 1 = x - 2 x — Ay = -8 y = 5 ,« -x + 4 2x + y = 6 y = x + 3 x- 2y y = 2 X + 6 x + 2y = —6 y = —3a; — . x — 3y = 6 y = 2x - 7 In Exercises 7-18, solve each of the given systems by sketching the lines represented by each equation of the given system on graph paper, then estimating the coordinates of the point of intersection to the nearest tenth. Check the solution. 10. 11. 12. —x - -3y = -- 3 x - -Ay = ---A Ax - -3y = = -12 —x - -4y = = 4 — 3x + 3y = -9 — 3x + 3y = -12 X - y = -2 2x- -2y = 6 Qx - -7y = -42 y = 1 -A X + Ax + 3y = 24 y = 1 -a; + 5 13. 14. 15. 16. 17. 18. 6x-7y= -42 y 7x — 8y = 56 y = x — A 1 6x + 3y = 12 -2a; - y = A x - Ay = -A -x + Ay = -A 3x + y = 3 -2x + 3y = -6 2x-y = -2 -x - 2y = A 260 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS In Exercises 19-24, use the graphing calculator to solve the given system. Round your answer to the nearest tenth. Use the Calculator Submission Guidelines from Chapter 3, Section 2, when reporting your answer on your homework. 19. 20. 21. y = -x + 7 4 y = -\x + 2 y = -x + 6 6 x + 3 y = —x — 3 J 3 4 v = —-X— 1 22. 23. 24. y = —x — 3 y 3 y y y y y --x-2 8 L -x + 1 ) --x + 5 -a; + 3 5 -x — 5 6 In Exercises 25-30, use the graphing calculator to solve the given system. Round your answer to the nearest tenth. Use the Calculator Submission Guidelines when reporting your answer on your homework. 25. 6x + 16y = 96 -6x + 13y = -78 26. -Ax + 16y = -64 5x + 8y = 40 27. -2z-llj/ = 22 8a: - 12y = -96 28. 29. 30. 6a; — lOy = 60 2a; - 18 j/ = -36 -6a; + 2y = -12 12a; + 3y = -36 — 3a; + y = -3 14a; + 3y = -42 j*- j*- ?*- Answers •** ~*^ •** 1.(4,-3) 3. (1,4) 5. (-2,-2) 7. (-3.4,0.1) 9. No solution. Lines are parallel. 11. (-1.8,4.5) 13. (-3.8,2.8) 15. No solution. Lines are parallel. 17. (1.4,-1.1) 19. (-4.6,3.5) 4.1. SOLVING SYSTEMS BY GRAPHING 261 21.(1.1,-1.6) 27. (-11.8,0.1) 23.(6.7,2.1) 29.(6.0,12.0) 25. (14.3,0.6) 262 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 4.2 Solving Systems by Substitution In this section we introduce an algebraic technique for solving systems of two equations in two unknowns called the substitution method. The substitution method is fairly straightforward to use. First, you solve either equation for either variable, then substitute the result into the other equation. The result is an equation in a single variable. Solve that equation, then substitute the result into any of the other equations to find the remaining unknown variable. You Try It! Solve the following system of EXAMPLE 1. Solve the following system of equations: equations: 2x-5y = -8 = 3x-l 9x + 2y V -19 13 + 3x V (4.7) (4.8) 3x- 1 5 2x — by (1,2 R c / / / / R Figure 4.27: 2x - 5y = -8 and y = x — 3 intersect at (1,2). Solution: Equation (4.8) is already solved for y. Substitute equation (4.8) into equation (4.7). This means we will substitute 3a: — 1 for y in equation (4.7). Equation (4.7). Substitute 3x — 1 for y in (4.7). Distribute —5. Simplify. Subtract 5 from both sides. Divide both sides by —13. As we saw in Solving Systems by Graphing, the solution to the system is the point of intersection of the two lines represented by the equations in the system. This means that we can substitute the answer x = 1 into either equation to find the corresponding value of y. We choose to substitute 1 for x in equation (4.8), then solve for y 7 but you will get exactly the same result if you substitute 1 for x in equation (4.7). 2x — 5y = -8 2x- 5(3x - 1) = -8 Now solve for X. 2x - - 15a; + 5 = -8 -13a; + 5 = -8 -13a; = -13 x = 1 3a; — 3(1) 2 Equation (4.8). Substitute 1 for x. Simplify. Hence, (x,y) = (1, 2) is the solution of the system. Check: To show that the solution (x,y) = (1,2) is a solution of the system, we need to show that (x,y) = (1,2) satisfies both equations (4.7) and (4.8). 4.2. SOLVING SYSTEMS BY SUBSTITUTION 263 Substitute (x,y) = (1,2) in equa- tion (4.7): 2x - by = -8 2(1) -5(2) = -8 2- 10 = -8 Thus, (1,2) satisfies equation (4.7). Substitute (x,y) = (1,2) in equa- tion (4.8): y = 3x — 1 2 = 3(1)- 1 2 = 3-1 2 = 2 Thus, (1,2) satisfies equation (4.8). Because (x,y) = (1,2) satisfies both equations, it is a solution of the system. Answer: (—3,4) Substitution method. The substitution method involves these steps: 1. Solve either equation for either variable. 2. Substitute the result from step one into the other equation. Solve the resulting equation. 3. Substitute the result from step two into either of the original system equa- tions or the resulting equation from step one (whichever seems easiest), then solve to find the remaining unknown variable. □ EXAMPLE 2. Solve the following system of equations: bx-2y = 12 Ax + y = 6 (4.9) (4.10) You Try It! Solve the following system of equations: x- 2y = 13 Ax - 3y = 26 Solution: The first step is to solve either equation for either variable. This means that we can solve the first equation for x or y, but it also means that we could first solve the second equation for x or y. Of these four possible choices, solving the second equation (4.10) for y seems the easiest way to start. 4a; + y = 6 y = 6 — Ax Equation (4.10). Subtract Ax from both sides. 264 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS 5x-2y = 12 5 J \ / \ 1 \ / \ / \ 1 - 5 ( 21/ \ 1 5 13, -18/12 )$ / \ / 1 \ 5- / \ Ax + y = 6 Figure 4.28: 5x - 2y = 12 and Ax + y = 6 intersect at (24/13,-18/13). Next, substitute 6 — Ax for y in equation (4.9). 5x-2y= 12 5x - 2(6 - Ax) = 12 bx - 12 + 8x = 12 13a;- 12 = 12 13a; = 24 _ 24 X ~ 13 Equation (4.9). Substitute 6 — 4a; for y in (4.9). Distribute —2. Simplify. Add 12 to both sides. Divide both sides by 13. Finally, to find the y-value, substitute 24/13 for x in the equation y = 6 — Ax (you can also substitute 24/13 for x in equations (4.9) or (4.10)). y = 6- Ax y = 6-4 /24 V13 y = 78 13 ~ 96 13 y = 18 ~13 Substitute 24/13 for x in y = 6 — Ax. Multiply, then make equivalent fractions. Simplify. Hence, (x, y) = (24/13, —18/13) is the solution of the system. Check: Let's use the graphing calculator to check the solution. First, we store 24/13 in X with the following keystrokes (see the result in Figure 4.29). 00BQ0 Next, we store —18/13 in the variable Y with the following keystrokes (see the result in Figure 4.29). EDQHHQS ALPHA □ ENTER Now, clear the calculator screen by pressing the CLEAR button, then enter the left-hand side of equation 4.9 with the following keystrokes (see the result in Figure 4.30). m X, T, 9, n m ALPHA □ ENTER Now enter the left-hand side of equation 4.10 with the following keystrokes (see the result in Figure 4.30). Note that each left-hand side produces the number on the right-hand sides of equations (4.9) and (4.10). Thus, the solution (x, y) = (24/13,-18/13) checks. 4.2. SOLVING SYSTEMS BY SUBSTITUTION 265 □ ALPHA □ ENTER 1.846153846 -18^13+V -1.384615385 Figure 4.29: Storing 24/13 and -18/13 in X and Y. 5*X-2*V 4+X+V 12 6 Figure 4.30: Checking equa- tions (4.9) and (4.10). Answer: (13/5,-26/5) □ EXAMPLE 3. Solve the following system of equations: 3x — 2y = 6 Ax + 5y = 20 You Try It! (4.11) (4.12) Solve the following system of equations: 3a; — 5y = 3 5x — 6y = 2 Solution: Dividing by —2 gives easier fractions to deal with than dividing by 3, 4, or 5, so let's start by solving equation (4.11) for y. 3x - -2y = 6 — 2y = 6 — 3a; 6 — 3a; y -2 y = -3 + -x Equation (4.11). Subtract 3x from both sides. Divide both sides by —2. Divide both 6 and —3a; by —2 using distributive property. 266 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS Substitute —3 + |x for y in equation (4.12). 4x + 5y = 20 f> (70/23,36/23)") -5 5 / / / / 5 3x - 2y = 6 Figure 4.31: 3x — 2y = 6 and 4x + 5y = 20 intersect at (70/23,36/23). 4x + 5 4x + 5y „ 3 —a; 2 20 20 Equation (4.12). Substitute -3 H — x for y. 2 4x- 15- 8x - 30 15 — x = 20 2 15x 23x 40 70 70 23 Distribute the 5. Clear fractions by multiplying both sides by 2. Simplify. Add 30 to both sides. Divide both sides by 23. To find y, substitute 70/23 for x into equation y You could also substitute 70/23 for x in equations (4.11) or (4.12) and get the same result. y = -3 + 3 -x v = 3/70 " 3 + U23 y = 69 105 ~23 + _ 23" y = 36 23 Substitute 70/23 for x. Multiply. Make equivalent fractions. Simplify. Hence, (x, y) = (70/23, 36/23) is the solution of the system. Check: To check this solution, let's use the graphing calculator to find the solution of the system. We already know that 3x — 2y = 6 is equivalent to y = —3 + |x. Let's also solve equation (4.12) for y. 4x 5y by y y 20 20 -Ax 20 -4.x , 4 4 x 5 Equation (4.12). Subtract 4x from both sides. Divide both sides by 5. Divide both 20 and — 4x by 5 using the distributive property. Enter y = — 3+ |x and y = 4— $x into the Y= menu of the graphing calculator (see Figure 4.32). Press the ZOOM button and select 6:ZStandard. Press 2ND CALC to open the CALCULATE menu, select 5:intersect, then press the ENTER key three times in succession to enter "Yes" to the queries "First curve," "Second curve," and "Guess." The result is shown in Figure 4.33. At the bottom of the viewing window in Figure 4.33, note how the coordinates of the point of intersection are stored in the variables X and Y. Without 4.2. SOLVING SYSTEMS BY SUBSTITUTION 267 Ploti Plots Plots WiB-3+3/2*X ^eB4-4^5*X ^7 = Figure 4.32: Enter y y = 4 tively. 3 + | a; and \x in Yl and Y2, respec- Intersection H=3.0H3H?B3 Y=i.Efi££lFH Figure 4.33: Use 5:intersect on the CALC menu to calculate the point of intersection. moving the cursor, (the variables X and Y contain the coordinates of the cursor), quit the viewing window by pressing 2ND QUIT, which is located above the MODE key. Then press the CLEAR button to clear the calculator screen. Now press the X,T,#,n key, then the MATH button on your calculator: This will open the MATH menu on your calculator (see Figure 4.34). Select l:^Frac, then press the ENTER key to produce the fractional equivalent of the decimal content of the variable X (see Figure 4.35). Hi U£ MUM CPX PRB Qe [►Frac 2 = ► Dec 3: 3 4: 3-K 5: *f 6'. fMinC 7-i-fMaxC Figure 4.34: The MATH menu. X^Frac V^Frac 70/23 36/23 Figure 4.35: Changing the contents of the variables X and Y to frac- tions. Repeat the procedure for the variable Y. Enter: ALPHA □ Select 1 : ►Frac, then press the ENTER key to produce the fractional equivalent of the decimal content of the variable Y (see Figure 4.35). Note that the fractional equivalents for X and Y are 70/23 and 36/23, precisely the same answers we got with the substitution method above. Answer: (-8/7,-9/7) □ 268 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS You Try It! Solve the following system of equations: x = —v 7 6x — 8y Figure 4.36: 2x + 3y = 6 and solution. 4 are parallel. No Answer: no solution Exceptional Cases Revisited It is entirely possible that you might apply the substitution method to a system of equations that either have an infinite number of solutions or no solutions at all. Let's see what happens should you do that. EXAMPLE 4. Solve the following system of equations: 2x + 3y = 6 2 y = — —x + 4 2 -— x 3 (4.13) (4.14) Solution: Equation (4.14) is already solved for y, so let's substitute for y in equation (4.13). 2x 2x + 3y = 6 Equation (4.13). 3 {-r +i ) = 6 2 Substitute — x 2x- 2x+ 12 = 6 Distribute the 3. 12 = 6 Simplify. 4 for y. Goodness! What happened to the x? How are we supposed to solve for x in this situation? However, note that the resulting statement, 12 = 6, is false, no matter what we use for x and y. This should give us a clue that there are no solutions. Perhaps we are dealing with parallel lines? Let's solve equation (4.13) for y, putting the equation into slope-intercept form, to help determine the situation. 2a;- 3y 3y: y 6 -2x- 2 --x Equation (4.13). Subtract 2x from both sides. Divide both sides by 3. Thus, our system is equivalent to the following two equations. 2 y = --x + 2 y = 2 — -x + 4 (4.15) (4.16) These lines have the same slope —2/3, but different y-intercepts (one has y- intercept (0,2), the other has y-intercept (0,4)). Hence, these are two distinct parallel lines and the system has no solution. □ 4.2. SOLVING SYSTEMS BY SUBSTITUTION 269 EXAMPLE 5. Solve the following system of equations: 2x-6y = -8 x = 3y — 4 Solution: Equation (4.18) is already solved for x, so let's substitute 3y — 4 for x in equation (4.17). You Try It! Solve the following system of (4.17) equations: (4.18) -2Sx+Uy = -126 A fnr y = 2x- 9 2x-6y = -8 2(3y - 4) - 6y = -8 6y-8-6y = -8 -8 = -8 Equation (4.17). Substitute 3y — 4 for x. Distribute the 2. Simplify. Goodness! What happened to the xl How are we supposed to solve for x in this situation? However, note that the resulting statement, —8 = —8, is a true statement this time. Perhaps this is an indication that we are dealing with the same line? Let's put both equations (4.17) and (4.18) into slope-intercept form so that we can compare them. Solve equation (4.17) for y: 2x Solve equation (4.18) for y: 6y = -8 6y = -2x- -8 -2a;- -8 V -e y = i 4 3 x - = 3y -4 x + 4 = = 3y x + 4 3 = y y = l = —X 3 4 + 3 Hence, the lines have the same slope and the same y-intercept and they are exactly the same lines. Thus, there are an infinite number of solutions. Indeed, any point on either line is a solution. Examples of solution points are (—4,0), (-1,1), and (2,2). y L ,i ■Vj 1 (- 1, L) (2 ,2 ("4, ') 5 5 _2x- 6« = - A s Figure 4.37: 2x - 6y = -8 and x = 3y — 4 are the same line. Infinite number of solu- tions. Answer: There are an infinite number of solutions. Examples of solution points are (0,-9), (5,1), and (-3,-15). □ Tip. When you substitute one equation into another and the variable disap- pears, consider: 1. If the resulting statement is false, then you have two distinct parallel lines and there is no solution. 2. If the resulting statement is true, then you have the same lines and there are an infinite number of solutions. 270 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS f> t* t* Exercises ■** •** ■** In Exercises 1-8, use the substitution method to solve each of the following systems. Check your answer manually, without the use of a calculator. 1. — 7a: + 7y = 63 5. x = —5 — 2y y = 6-2x -2x-Gy = 18 2. 3a;-8y = 27 6. x = 15 + 6y y = 4 - 7x 9x + 3y = 21 3. x = 19 + 7y 7. 6a; - 8y = 24 3a: — 3y = 3 y = 15 + 3a; 4. a; = 39 + 8y 8. 9a; + 8y = -45 -9a; + 2y = -71 y = 15 - 8a; In Exercises 9-28, use the substitution method to solve each of the following systems. 18. -6x-6y = 102 9x — y = —63 10. -a; + 9y = -12 19. -Ax - 8y = -4 y = -2j/-4 11. -x + 4y = 22 20. 3x + 6y = 2 y = -2y + 2 12. -x-2y=15 21. -2a;-2y = 26 -7a; + y = 19 13. x + 2y = -A 22. -2x - 8y = -30 — x + 9y = = 46 7x — Ay = = -27 —x + 9y = = -12 Ax + 7y = = -38 —x + Ay = = 22 8x + 7y = = -20 -x-2y = = 15 3a; — 9y = = 15 x + 2y = -4 6x — Ay = -56 x + 8y = 79 Ax + Qy = 82 x + 6y = -49 -3x + Ay = -7 x — Ay = 33 Ax + 7y = -6 -2x + 8y = -50 —9x — y = -3 —6a; + y = 10 14. a; + 8y = 79 23. 3a; - Ay = -43 -3x + y = 22 15. a; + 6y=-49 24. -2x + 8y = 14 8x + y = A3 16. a;-4y = 33 25. -8a; - Ay = 24 9x-y = -71 17. -2a; + 8y=-50 26. -8x - 2y = -1A —6a; — y = —9 4.2. SOLVING SYSTEMS BY SUBSTITUTION 271 27. -8x-7y = 2 28. 9x + 4y = -3 8 9 y = — —x + 9 y = ——x + o In Exercises 29-36, use the substitution method to solve each of the following systems. Use your graphing calculator to check your solution. 29. 3x-5y = 3 33. 3x + 8y = 6 5x - 7y = 2 2x + 7y = -2 30. 4a; - 5y = 4 34. 3a; - 7y = 6 3a; — 2y = — 1 2a; — 3y = 1 31. 4x + 3y = 8 35. 4a; + 5y = 4 3a; + 4 y = 2 -3x - 2y = 1 32. 3a; + 8y = 3 36. 5a; + Ay = 5 -4a; - 9y = -2 4a; + 5y = 2 In Exercises 37-48, use the substitution method to determine how many solutions each of the following linear systems has. 37. -9a; + 6y = 9 43. y = 7y + 18 9a; - 63y = 162 44. y = Ay - 9 -10a; + 4Ch/ = 90 39. 40. -9a; + 6y = 9 3 V = —x — 8 y 2 3a; — 5y = 9 3 ^fi u = -x + 5 y = -2a;- 16 -14a;- 7y = 112 y = -12a;- f 12 120a;+102/= 120 x = 16 — 5y -Ax + 2y = 2A x = -18- Ay 7x-7y = 49 45. 46. a;= -2y + 3 Ax + 8y = A x = 2y + A —3x + 6y = 5 41. a;=16-5w 47 - -9a; + 4y = 73 y = —3 — 2x 42. a;=-18-4y 48. 6x + 9y = 27 y = 16 — 5a; 1. -1,8) 3. -2,-3) 5. 3,-4) 7. -8,-9) 9. -1,5) 11. -6,4) 13. -8,2) 15. -7,-7) 17. 1,-6) 19. 21. -4,-9) 23. -5,7) 272 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS £»- j*- &*■ Answers •*$ **s ■*$ 25. (-7,8) 27. No solution 29. (-11/4,-9/4) 31. (26/7,-16/7) 33. (58/5,-18/5) 35. (-13/7,16/7) 37. No solutions 39. Infinite number of solutions 41. One solution 43. Infinite number of solutions 45. No solutions 47. One solution 4.3. SOLVING SYSTEMS BY ELIMINATION 273 4.3 Solving Systems by Elimination When both equations of a system are in standard form Ax + By = C, then a process called elimination is usually the best procedure to use to find the solution of the system. Elimination is based on two simple ideas, the first of which should be familiar. 1. Multiplying both sides of an equation by a non-zero number does not change its solutions. Thus, the equation x + 3y = 7 (4.19) will have the same solutions (it's the same line) as the equation obtained by multiplying equation (4.19) by 2. 2x + 6y = 14 (4.20) 2. Adding two true equations produces another true equation. For example, consider what happens when you add 4 = 4 to 5 = 5. Even more importantly, consider what happens when you add two equa- tions that have (2, 1) as a solution. The result is a third equation whose graph also passes through the solution. X + V = 3 X - y = 1 2x = 4 X = 2 > X Fact. Adding a multiple of an equation to a second equation produces and equation that passes through the same solution as the first two equations. 274 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS One more important thing to notice is the fact that when we added the equations x + y = 3 x - y = I 2x =4 the variable y was eliminated. This is where the elimination method gets its name. The strategy is to somehow add the equations of a system with the intent of eliminating one of the unknown variables. However, sometimes you need to do a little bit more than simply add the equations. Let's look at an example. You Try It! Solve the following system of EXAMPLE 1. Solve the following system of equations, equations: x + 2y = -5 2x — y = —5 x + 3y = 14 -8x-3y = -28 (4.21) (4.22) Solution: Our focus will be on eliminating the variable x. Note that if we multiply equation (4.21) by —2, then add the result to equation (4.22), the x terms will be eliminated. -2x 2x v 10 -5 % Multiply equation (4.21) by -2. Equation (4.22). Add the equations. I r > '2s- V = -5 7 / / / / £ 7 k -3 j 1) c / / / 5- Divide both sides of — by = 5 by — 5 to get y = — 1. To find the corresponding value of x, substitute —1 for y in equation (4.21) (or equation (4.22)) and solve for x. Equation (4.21) Substitute —1 for y. Solve for x. x + 2y = — 5 Check: To check, we need to show that the point (x, y) = (—3,1) satisfies both equations. Figure 4.38: x + 2y = -5 and 2x - y = -5 intersect at Substitute (x,y) = (-3,-1) into Substitute (x,y) = (-3,-1) into ( — 3,— 1). equation (4.21). equation (4.22). x + 2y = -5 + 2(-l) = -5 x = -3 x + 2y -3 + 2(-l) -5 2x- -y = -5 2(" -3) "(- -5 = -5 -5 Answer: (2,4). Thus, the point (x, y) = (—3, —1) satisfies both equations and is therefore the solution of the system. □ 4.3. SOLVING SYSTEMS BY ELIMINATION 275 To show that you have the option of which variable you choose to eliminate, let's try Example 1 a second time, this time eliminating y instead of x. EXAMPLE 2. Solve the following system of equations. x + 2y = -5 (4.23) 2x-y=-5 (4.24) Solution: This time we focus on eliminating the variable y. We note that if we multiply equation (4.24) by 2, then add the result to equation (4.23), the y terms will be eliminated. x Ax 2y 2y 5.7,' -5 Equation (4.23). -10 Multiply equation (4.24) by 2. -15 Add the equations. Divide both sides of 5a; = —15 by 5 to get x = — 3. To find the corresponding value of y, substitute —3 for x in equation (4.23) (or equation (4.24)) and solve for y. x -3 •2» 2y 2y y Equation (4.23) Substitute —3 for x. Add 3 to both sides. Divide both sides by 2. Hence, (x, y) = (—3, —1), just as in Example 1, is the solution of the system. Sometimes elimination requires a thought process similar to that of finding a common denominator. EXAMPLE 3. Solve the following system of equations. 3x + Ay = 12 2x - 5y = 10 (4.25) (4.26) Solution: Let's focus on eliminating the x-terms. Note that if we multiply equation (4.25) by 2, then multiply equation (4.26) by —3, the x-terms will be eliminated when we add the resulting equations. 6.7' -Qx 8y 15y 24 -30 23y Multiply equation (4.25) by 2. Multiply equation (4.26) by —3. Add the equations. You Try It! Figure 4.39: x + 2y = -5 and 2x — y = —5 intersect at (-3,-1). □ You Try It! Solve the following system of equations: -I4x + 9y 7x + 3y 94 -62 276 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS Hence, y = -6/23. At this point, we could substitute y = —6/23 in either equation, then solve the result for x. However, working with y = —6/23 is a bit daunting, partic- ularly in the light of elimination being easier. So let's use elimination again, this time focusing on eliminating y. Note that if we multiply equation (4.25) by 5, then multiply equation (4.26) by 4, when we add the results, the y-terms will be eliminated. 12 Figure 4.40: 3a; + 4y = 12 and 2x — 5y = 10 intersect at (100/23,-6/23). Answer: 15a; 8x 20y 20y 60 40 23a: Thus, x Check: Multiply equation (4.25) by 5. Multiply equation (4.26) by 4. = 100 Add the equations. 100/23, and the system of the system is (x,y) = (100/23, -6/23). Let's use the graphing calculator to check the solution. First, store 100/23 in X, then —6/23 in Y (see Figure 4.41). Next, enter the left-hand sides of equations (4.25) and (4.26). 4.347826087 -6^23+V -.2608695652 3*X+4*V 2*X-5*Y 12 10 Figure 4.41: Enter 100/23 in X, -6/23 in Y. Figure 4.42: Enter the left-hand sides of equations (4.25) and (4.26). Note that both calculations in Figure 4.42 provide the correct right-hand sides for equations (4.25) and (4.26). Thus, the solution (x,y) = (100/23,-6/23) checks. □ You Try It! Exceptional Cases In the previous section, we saw that if the substitution method led to a false statement, then we have parallel lines. The same thing can happen with the elimination method of this section. Solve the following system of EXAMPLE 4. Solve the following system of equations, equations: 5a; — Ay 15a; - 12y 16 x + y 2x + 2y -6 (4.27) (4.28) 49 4.3. SOLVING SYSTEMS BY ELIMINATION 277 Solution: Let's focus on eliminating the x-terms. Note that if we multiply equation (4.27) by —2, the x-terms will be eliminated when we add the resulting equations. -2x - 2y = -6 Multiply equation (4.27) by -2. 2x + 2y = -6 Equation (4.28). = —12 Add the equations. Because of our experience with this solving this exceptional case with substi- tution, the fact that both variables have disappeared should not be completely surprising. Note that this last statement is false, regardless of the values of x and y. Hence, the system has no solution. Indeed, if we find the intercepts of each equation and plot them, then we can easily see that the lines of this system are parallel (see Figure 4.43). Parallel lines never intersect, so the system has no solutions. x + y = 3 2x + 2y = -6 Figure 4.43: The lines x + y = 3 and 2x + 2y = —6 are parallel. Answer: no solution In the previous section, we saw that if the substitution method led to a true statement, then we have the same lines. The same thing can happen with the elimination method of this section. □ EXAMPLE 5. Solve the following system of equations. x — 7y = 4 -3x + 2ly = -12 You Try It! Solve t le following sys ;em of (4.29) equations: (4.30) 2x - ly = 4 8a; - 28y = 16 278 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS Solution: If we are not on automatic pilot late at night doing our homework, we might recognize that the equations 4.29 and (4.30) are identical. But it's also conceivable that we don't see that right away and begin the elimination 4 method. Let's multiply the first equation by 3, then add. This will eliminate the x-terms. 3a: - 21y = 12 Multiply equation (4.29) by 3. -3a; + 21y = -12 Equation (4.30). = Add the equations. Fieure 4 44- x — 7v = 4 and Again, all of the variables have disappeared! However, this time the last state- o„ , oi v — _io are the men t is true, regardless of the values of x and y. same line Infinite number of Notice that if we multiply equation (4.29) by —3, then we have two identical solutions. equations. Answer: There are an infinite number of solutions. Examples of solution points are (2,0), (9,2), and (-5,-2). -3a; + 2ly = -12 Multiply equation (4.29) by 3. -3a; + 21y = -12 Equation (4.30). The equations are identical! Hence, there are an infinite number of points of intersection. Indeed, any point on either line is a solution. Example points of solution are (-3,-1), (0,-4/7), and (4,0). □ 4.3. SOLVING SYSTEMS BY ELIMINATION 279 ti. ;». ;». Exercises ■*j ■*-: •*; In Exercises 1-8, use the elimination method to solve each of the following systems. Check your result manually, without the assistance of a calculator. x + Ay 9x 7.y -43 x + 6y = -53 5a; — 9y = 47 6a; + 2/ = 8 4a; + 2y = Ax + y = 18 — 2x + 6y = = -22 7. -8a; + y Ax + 3y = -56 56 2x + y = 7x + 8y = = 21 = 87 x + 8y — bx — 9y 41 = -50 x — Ay —2x — 6y = -31 = -36 In Exercises 9-16, use the elimination method to solve each of the following systems. 10. 11. 12. -12a; + 9y = —6a; — Ay = -34 -27a; - by = 148 — 9x — 3y = 60 27a; — 6y = -96 -3x — by = 22 -8x + 8y = -32 2x - 9y = 15 13. 14. 15. 16. 2x - Qy 28 3x + I8y = -60 8x — 6y 96 Ax + 30y = -156 32a; + 111 = -238 8a: - Mj 64 2x + Qy = 30 2x + 7y = 51 In Exercises 17-24, use the elimination method to solve each of the following systems. 17. 18. 19. 20. 3.r - 7y = -75 -2a; - 2y = -10 -8a; + 3y = 42 -7x + 8y = 20 9x - - 9y = -63 2x - - Qy = -34 -Ax - 8y = -52 -7x - 3y = -14 21. 22. 23. 24. 9;r - 2y = 28 5 a: - 3y = = -32 8a; — 2y = = -12 Gx + 3y = 12 3x — by = = -34 7x + 7y ~- 56 9x — 9y = = 9 7x + Ay = = 8 280 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS In Exercises 25-32, use the elimination method to solve each of the following systems. Use your calculator to check your solutions. 25. 26. 27. 28. 2a; - 7x + 7y = 6y = -2 3 -9x - bx — Ay = 3y = 4 -1 2x + -5x + 3y = by = -2 2 -bx + -Ax - 8y = 7y = -3 3 29. 30. 31. 32. 9a; + Ay — 7x — 9y = -A 3 — 3a; — by Ax + 6y = -4 = 1 2x + 2y = 3a; — by - = 4 = 3 6a; — 9y -Ax - %y = -2 4 In Exercises 33-40, use the elimination method to determine how many solutions each of the following system of equations has. 33. 34. 35. 36. X + 7y = -32 -8a; - 56y = 256 -8a; + y = -53 56a; - 7y = 371 16a; — 16i/ = -256 -8a; + 8y = 128 3.r — 3y = 42 -6a; + 6y = -84 37. 38. 39. 40. X - Ay = -37 2x - 8y = 54 Ax + y = -13 28a; + 7y = 189 X + 9y = 73 -Ax - by = -44 O.r + y = 31 — 5a; - 6y = -62 j*- j*- ?*- Answers •** ~*^ ■** 1. (-4,1) 3. (2,-4) 5. (8,8) 7.(1,5) 9. (3,4) 11. (-4,-2) 13. (8,-2) 15. (7,-2) 17. (-4,9) 19. (-2,5) 21. (-4,4) 23. (3,5) 4.3. SOLVING SYSTEMS BY ELIMINATION 281 25. (9/61, 20/61) 33. Infinite number of solutions 27. (—16/25, —6/25) 35. Infinite number of solutions 29. (-24/53, 1/53) 37. No solutions 31. (13/8,3/8) 39. One solution 282 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS You Try It! If the second of two complementary angles is 6 degrees larger than 3 times the first angle, find the degree measure of both angles. 4.4 Applications of Linear Systems In this section we create and solve applications that lead to systems of linear equations. As we create and solve our models, we'll follow the Requirements for Word Problem Solutions from Chapter 2, Section 5. However, instead of setting up a single equation, we set up a system of equations for each application. EXAMPLE 1. In geometry, two angles that sum to 90° are called comple- mentary angles. If the second of two complementary angles is 30 degrees larger than twice the first angle, find the degree measure of both angles. Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. Our variable dictionary will take the form of a diagram, naming the two complementary angles a and /3. 2. Set up a Systems of Equations. The "second angle is 30 degrees larger than twice the first angle" becomes P = 30 + 2a (4.31) Secondly, the angles are complementary, meaning that the sum of the angles is 90°. a + /3 = 90 (4.32) Thus, we have a system of two equations in two unknowns a and (3. 3. Solve the System. As equation (4.31) is already solved for /?, let use the substitution method and substitute 30 + 2a for j3 in equation (4.32). a + (3 = 90 a + (30 + 2a) = 90 3a + 30 = 90 3a = 60 a = 20 Equation (4.32). Substitute 30 + 2a for (3. Combine like terms. Subtract 30 from both sides. Divide both sides by 3. 4.4. APPLICATIONS OF LINEAR SYSTEMS 283 4. Answer the Question. The first angle is a = 20 degrees. The second angle is: f3 = 30 + 2a /3 = 30 + 2(20) /3 = 70 Equation (4.31). Substitute 20 for a. Simplify. 5. Look Back. Certainly 70° is 30° larger than twice 20°. Also, note that 20° + 70° = 90°, so the angles are complementary. We have the correct solution. Answer: 21 and 69 □ EXAMPLE 2. The perimeter of a rectangle is 280 feet. The length of the rectangle is 10 feet less than twice the width. Find the width and length of the rectangle. Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. Our variable dictionary will take the form of a diagram, naming the width and length W and L, respectively. You Try It! The perimeter of a rectangle is 368 meters. The length of the rectangle is 34 meters more than twice the width. Find the width and length of the rectangle. W L W 2. Set up a System of Equations. The perimeter is found by summing the four sides of the rectangle. P=L+W +L+W P = 2L + 2W We're told the perimeter is 280 feet, so we can substitute 280 for P in the last equation. 280 = 2L + 2W 284 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS We can simplify this equation by dividing both sides by 2, giving the following result: L + W = 140 Secondly, we're told that the "length is 10 feet less than twice the width." This translates to: L = 2W - 10 Thus, the system we need to solve is: L + W = U0 (4.33) L = 2W-W (4.34) 3. Solve the System. As equation (4.34) is already solved for L, let use the substitution method and substitute 2W — 10 for L in equation (4.33). W + L = 140 Equation (4.33). W + (2W - 10) = 140 Substitute 2W - 10 for L. 3W - 10 = 140 Combine like terms. 3W = 150 Add 10 to both sides. W = 50 Divide both sides by 3. 4. Answer the Question. The width is W = 50 feet. The length is: L = 2W-W Equation (4.34). L = 2(50) - 10 Substitute 50 for W. L = 90 Simplify. Thus, the length is L = 90 feet. 5. Look Back. Perhaps a picture, labeled with our answers might best demonstrate that we have the correct solution. Remember, we found that the width was 50 feet and the length was 90 feet. 90 50 50 Answer: length = 134, width = 50 90 Note that the perimeter is P = 90 + 50 + 90 + 50 = 280 feet. Secondly, note that the length 90 feet is 10 feet less than twice the width. So we have the correct solution. □ 4.4. APPLICATIONS OF LINEAR SYSTEMS 285 You Try It! EXAMPLE 3. Pascal has $3.25 in change in his pocket, all in dimes and Eloise has $7.10 in change in quarters. He has 22 coins in all. How many dimes does he have? her pocket, all in nickels and quarters, she has 46 coins in all. How many quarters does she have? Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1 . Set up a Variable Dictionary. Let D represent the number of dimes and let Q represent the number of quarters. 2. Set up a System of Equations. Using a table to summarize information is a good strategy. In the first column, we list the type of coin. The second column gives the number of each type of coin, and the third column contains the value (in cents) of the number of coins in Pascal's pocket. Number of Coins Value (in cents) Dimes D WD Quarters Q 25Q Totals 22 325 Note that D times, valued at 10 cents apiece, are worth 10D cents. Sim- ilarly, Q quarters, valued at 25 cents apiece, are worth 25Q cents. Note also how we've change $3.25 to 325 cents. The second column of the table gives us our first equation. D + Q = 22 (4.35) The third column of the table gives us our second equation. 10£> + 25Q = 325 (4.36) Solve the System. Because equations (4.35) and (4.36) are both in stan- dard form Ax + By = C, we'll use the elimination method to find a solution. Because the question asks us to find the number of dimes in Pascal's pocket, we'll focus on eliminating the Q-terms and keeping the D-terms. -25D 10D 25Q 25Q -15D -550 Multiply equation (4.35) by -25. 325 Equation (4.36). -225 Add the equations. Dividing both sides of the last equation by —15 gives us D = 15. 4. Answer the Question. The previous solution tells us that Pascal has 15 dimes in his pocket. 286 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS Answer: 24 5. Look Back. Again, summarizing results in a table might help us see if we have the correct solution. First, because we're told that Pascal has 22 coins in all, and we found that he had 15 dimes, this means that he must have 7 quarters. Number of Coins Value (in cents) Dimes 15 150 Quarters 7 175 Totals 22 325 Fifteen dimes are worth 150 cents, and 7 quarters are worth 175 cents. That's a total of 22 coins and 325 cents, or $3.25. Thus we have the correct solution. □ You Try It! Eileen inherits $40,000 and decides to invest the money in two accounts, part in a certificate of deposit that pays 3% interest per year, and the rest in a mutual fund that pays 6% per year. At the end of the first year, her investments earn a total of $2,010 in interest. Find the amount invested in each account. EXAMPLE 4. Rosa inherits $10,000 and decides to invest the money in two accounts, one portion in a certificate of deposit that pays 4% interest per year, and the rest in a mutual fund that pays 5% per year. At the end of the first year, Rosa's investments earn a total of $420 in interest. Find the amount invested in each account. Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. Let C represent the amount invested in the certificate of deposit and M represent the amount invested in the mutual fund. 2. Set up a System of Equations. We'll again use a table to summarize information. Rate Amount invested Interest Certificate of Deposit 4% C 0.04C Mutual Fund 5% M 0.05M Totals 10,000 420 At 4%, the interest earned on a C dollars investment is found by taking 4% of C (i.e., 0.04C). Similarly, the interest earned on the mutual fund is 0.05M. 4.4. APPLICATIONS OF LINEAR SYSTEMS 287 The third column of the table gives us our first equation. The total investment is $10,000. C + M = 10000 The fourth column of the table gives us our second equation. The total interest earned is the sum of the interest earned in each account. 0.04(7 + 0.05M = 420 Let's clear the decimals from the last equation by multiplying both sides of the equation by 100. 4(7 + 5M = 42000 Thus, the system we need to solve is: C + M = 10000 (4.37) 4(7 + 5M = 42000 (4.38) 3. Solve the System. Because equations (4.37) and (4.38) are both in stan- dard form Ax + By = C, we'll use the elimination method to find a solution. We'll focus on eliminating the (7-terms. -AC - AM = -40000 Multiply equation (4.37) by -4. AC + 5M = 42000 Equation (4.38). M = 2000 Add the equations. Thus, the amount invested in the mutual fund in M = $2, 000. 4. Answer the Question. The question asks us to find the amount invested in each account. So, substitute M = 2000 in equation (4.37) and solve for (7. C + M = 10000 Equation (4.37). C + 2000 = 10000 Substitute 2000 for M. C = 8000 Subtract 2000 from both sides. Thus C = $8, 000 was invested in the certificate of deposit. 5. Look Back. First, note that the investments in the certificate of deposit and the mutual fund, $8,000 and $2,000 respectively, total $10,000. Let's calculate the interest on each investment: 4% of $8,000 is $320 and 5% of $2,000 is $100. Rate Amount invested Interest Certificate of Deposit 4% 8,000 320 Mutual Fund 5% 2, 000 100 Totals 10,000 420 288 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS Answer: $13,000 in the certificate of deposit, $27,000 in the mutual fund. Note that the total interest is $420, as required in the problem statement. Thus, our solution is correct. □ You Try It! A store sells peanuts for $4.00 per pound and pecans for $7.00 per pound. How many pounds of peanuts and how many pounds of pecans should you mix to make a 25-lb mixture costing $5.80 per pound? EXAMPLE 5. Peanuts retail at $0.50 per pound and cashews cost $1.25 per pound. If you were a shop owner, how many pounds of peanuts and cashews should you mix to make 50 pounds of a peanut-cashew mixture costing $0.95 per pound? Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. Let P be the number of pounds of peanuts used and let C be the number of pounds of cashews used. 2. Set up a System of Equations. We'll again use a table to summarize information. Cost per pound Amount (pounds) Cost Peanuts $0.50 P 0.50P Cashews $1.25 C 1.25C Totals $0.95 50 0.95(50)=47.50 At $0.50 per pound, P pounds of peanuts cost 0.50P. At $1.25 per pound, C pounds of cashews cost 1.25C. Finally, at $0.95 per pound, 50 pounds of a mixture of peanuts and cashews will cost 0.95(50), or $47.50. The third column of the table gives us our first equation. The total number of pounds of mixture is given by the following equation: P + C = 50 The fourth column of the table gives us our second equation. The total cost is the sum of the costs for purchasing the peanuts and cashews. 0.50P + 1.25C = 47.50 Let's clear the decimals from the last equation by multiplying both sides of the equation by 100. 50P + 125C = 4750 Thus, the system we need to solve is: P + C = 50 (4.39) 50P + 125C = 4750 (4.40) 4.4. APPLICATIONS OF LINEAR SYSTEMS 289 3. Solve the System. Because equations (4.39) and (4.40) are both in stan- dard form Ax + By = C, we'll use the elimination method to find a solution. We'll focus on eliminating the P-terms. -50P 50P 50C 125C* -2500 4750 Multiply equation (4.39) by -50. Equation (4.40). 75 C = 2250 Add the equations. Divide both sides by 75 to get C = 30 pounds of cashews are in the mix. Answer the Question. The question asks for both amounts, peanuts and cashews. Substitute C = 30 in equation (4.39) to determine P. P P - -C 30 P 50 50 20 Equation (4.39). Substitute 30 for C. Subtract 30 from both sides. Thus, there are P = 20 pounds of peanuts in the mix. Look Back. First, note that the amount of peanuts and cashews in the mix is 20 and 30 pounds respectively, so the total mixture weighs 50 pounds as required. Let's calculate the costs: for the peanuts, 0.50(20), or $10, for the cashews, 1.25(30) = 37.50. Cost per pound Amount (pounds) Cost Peanuts $0.50 20 $10.00 Cashews $1.25 30 $37.50 Totals $0.95 50 47.50 Note that the total cost is $47.50, as required in the problem statement. Thus, our solution is correct. Answer: 10 pounds of peanuts, 15 pounds of pecans □ 290 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS f»- n- fa- Exercises ■*1 ■*'» .*5 1. In geometry, two angles that sum to 90° are called complementary angles. If the second of two complementary angles is 42 degrees larger than 3 times the first angle, find the degree measure of both angles. 2. In geometry, two angles that sum to 90° are called complementary angles. If the second of two complementary angles is 57 degrees larger than 2 times the first angle, find the degree measure of both angles. 3. The perimeter of a rectangle is 116 inches. The length of the rectangle is 28 inches more than twice the width. Find the width and length of the rectangle. 4. The perimeter of a rectangle is 528 inches. The length of the rectangle is 24 inches more than twice the width. Find the width and length of the rectangle. 5. Maria has $6.35 in change in her pocket, all in nickels and quarters, she has 59 coins in all. How many quarters does she have? 6. Amy has $5.05 in change in her pocket, all in nickels and quarters, she has 53 coins in all. How many quarters does she have? 7. A store sells cashews for $6.00 per pound and raisins for $7.00 per pound. How many pounds of cashews and how many pounds of raisins should you mix to make a 50-lb mixture costing $6.42 per pound? 8. A store sells cashews for $3.00 per pound and pecans for $8.00 per pound. How many pounds of cashews and how many pounds of pecans should you mix to make a 50-lb mixture costing $4.10 per pound? 9. Roberto has $5.45 in change in his pocket, all in dimes and quarters, he has 38 coins in all. How many dimes does he have? 10. Benjamin has $7.40 in change in his pocket, all in dimes and quarters, he has 44 coins in all. How many dimes does he have? 11. In geometry, two angles that sum to 180° are called supplementary angles. If the second of two supplementary angles is 40 degrees larger than 3 times the first angle, find the degree measure of both angles. 12. In geometry, two angles that sum to 180° are called supplementary angles. If the second of two supplementary angles is 114 degrees larger than 2 times the first angle, find the degree measure of both angles. 13. Eileen inherits $20,000 and decides to in- vest the money in two accounts, part in a certificate of deposit that pays 3% in- terest per year, and the rest in a mutual fund that pays 5% per year. At the end of the first year, her investments earn a total of $780 in interest. Find the amount invested in each account. 14. Alice inherits $40,000 and decides to in- vest the money in two accounts, part in a certificate of deposit that pays 3% in- terest per year, and the rest in a mutual fund that pays 6% per year. At the end of the first year, her investments earn a to- tal of $1,980 in interest. Find the amount invested in each account. 15. The perimeter of a rectangle is 376 cen- timeters. The length of the rectangle is 12 centimeters less than three times the width. Find the width and length of the rectangle. 16. The perimeter of a rectangle is 344 feet. The length of the rectangle is 28 feet less than three times the width. Find the width and length of the rectangle. 4.4. APPLICATIONS OF LINEAR SYSTEMS 291 $*•$*<$*■ Answers •*$ *** •** 1. 12° and 78° 3. Length is 48 inches, width is 10 inches 5. 17 quarters 7. 29 pounds of cashews, 21 pounds of raisins 9. 27 dimes 11. 35° and 145° 13. $11,000 in certificate of deposit, $9,000 in mutual fund. 15. Length is 138 centimeters, width is 50 cen- timeters Chapter 5 Polynomial Functions Polynomials appear in a wide variety of applications in mathematics and sci- ence. Solving algebraic equations containing polynomials is among the oldest problems in mathematics. The Babylonians solved quadratic equations as early as 2000 BC. Euclid of Alexandria used geometry to solve quadratic equations in 300 BC. Arab mathematicians solved quadratic equations in 1000 AD. The current method of solving equations using more modern notation started in 1557 when Robert Recorde used the equal sign in The Whetstone of Witte. In this chapter, we will learn about the properties of polynomials, as well as how to manipulate and use them in application problems. 293 294 CHAPTER 5. POLYNOMIAL FUNCTIONS 5.1 Functions We begin with the definition of a relation. Relation. A relation is a collection of ordered pairs. The collection of ordered pairs R = {(0,3), (0,4), (1,5), (2, 6)} is an example of a relation. If we collect the first element of each ordered pair in a set, we have what is called the domain of the relation. Domain. The domain of a relation is the set of all first elements of the ordered pairs. For example, in the relation R = {(0, 3), (0, 4), (1, 5), (2, 6)}, if we collect the first element of each ordered pair in i?, we get the domain: Domain of R = {0,1,2} Although the number zero appears twice as a first element in the ordered pairs of R, note that we list it only once when listing the elements in the domain of R. In similar fashion, if we collect the second elements of each ordered pair in a set, we have what is called the range of a relation. Range. The range of a relation is the set of all second elements of the ordered pairs. For example, in the relation R = {(0, 3), (0, 4), (1, 5), (2, 6)}, if we collect the second element of each ordered pair in i?, we get the range: Range of R= {3,4,5,6} You Try It! State the domain and range EXAMPLE 1. State the domain and range of the relation of the relation T = {(5,3), (6,3), (7,4)}. S = {(-1,7), (2,5), (2,3)}. Solution: Collect the first element of each ordered pair in T to list the domain: Domain of T = {5, 6, 7} 5.1. FUNCTIONS 295 Collect the second element of each ordered pair in T to list the range: Range of T = {3, 4} Note that even though the number three appears in the second position twice, Answer: we list it only once in describing the range. Domain of S = { — 1, 2}, Range of S = {3, 5, 7} □ You Try It! EXAMPLE 2. State the domain and range of the relation shown in Figure 5.1. State the domain and range of the relation shown below. y y C B D ■*■ x 5' L J 5 *, A 5 > ■ h Figure 5.1: We can present a relation as a collection of ordered pairs in a graph. Solution: Point A has coordinates (3,2), point B has coordinates (—2,3), point C has coordinates (—4,-3), and point D has coordinates (1,-3). We can collect these points in a set. 5 = {(3,2), (-2,3), (-4,-3), (1,-3)} If we collect each element in the first position of each ordered pair, we have the domain. Domain of S = {-4, -2, 1,3} Note that it is traditional to list the domain elements in order (smallest to largest). Next, if we collect each element in the second position of each ordered pair, we have the range. Range of S= {-3,2,3} 296 CHAPTER 5. POLYNOMIAL FUNCTIONS Answer: Domain of 5 = {-2,1,2}, Range of 5 = {-4,-1,2,3} Again, it is traditional to list the elements in order. Note again that we did not repeat the number —3 in listing the range, even though it is used twice as a second element of an ordered pair in the set S. □ Mapping Diagrams A mapping diagram is a useful construction that helps one to analyze a relation. Consider the earlier relation R = {(0, 3), (0, 4), (1, 5), (2, 6)}, which had domain V = {0, 1, 2} and range TZ = {3, 4, 5, 6}. To construct a mapping diagram for R, list the elements in the domain of R on the left, list the elements of the range of R on the right, then use arrows to indicate the ordered pairs (see Figure 5.2). R Figure 5.2: Using a mapping diagram to describe the relation R. Note how the ordred pair (0, 3) is indicated by drawing an arrow connecting on the left to 3 on the right. We say that the relation "maps to 3" and write R : — > 3. In similar fashion: • The ordered pair (0, 4) is indicated by drawing an arrow connecting on the left to 4 on the right; i.e., R "maps to 4" or R : — > 4. • The ordered pair (1,5) is indicated by drawing an arrow connecting 1 on the left to 5 on the right; i.e., R "maps 1 to 5" or R : 1 — > 5. • The ordered pair (2, 6) is indicated by drawing an arrow connecting 2 on the left to 6 on the right; i.e., R "maps 2 to 6" or R : 2 — > 6. You Try It! EXAMPLE 3. Create a mapping diagram for the relation in Example 1. Solution. The relation of Example 1 is T = {(5, 3), (6, 3), (7, 4)}. List the domain T> = {5, 6, 7} on the left, the range TZ = {3, 4} on the right, then use arrows to indicate the ordered pairs (see Figure 5.3). □ 5.1. FUNCTIONS 297 Figure 5.3: A mapping diagram for the relation T. Function Definition A function is a very special type of relation. Function A relation is a function if and only if each object in the domain is paired with exactly one object in the range. As a first example, consider the relation R = {(0,3), (0,4), (1,5), (2,6)} whose mapping diagram is pictured in Figure 5.2. Note that in the domain is paired with two objects, 3 and 4, in the range. Hence, relation R is is not a function. As a second example, consider the relation T = {(5, 3), (6, 3), (7, 4)}, whose mapping diagram is pictured in Figure 5.3. In this example, each domain object is paired with exactly one range object: 5 only gets sent to 3, 6 only gets sent to 3, and 7 only gets sent to 4. Hence, the relation T is a function. The fact that the range object 3 is used twice does not matter. It's the fact that each domain object gets sent to exactly one range object that matters. You Try It! EXAMPLE 4. Consider the relation pictured in Figure 5.4. Is it a function? Consider the relation pictured below. Is it a y function? . , E D B A C 1 L i- 5 1 / c ^ 4 5 r n R r J ■ -* X Figure 5.4: Is this relation a function? 298 CHAPTER 5. POLYNOMIAL FUNCTIONS Solution: The graph in Figure 5.4 consists of the points .4(1,1.), 5(2,2), C(3,2), £>(3,3), and £'(4,4). The domain is V = {1,2,3,4} and the range is 1Z = {1,2,3,4}. A mapping diagram (see Figure 5.5) will help us decide if the relation represented by the graph is a function. Put the domain on the left, the range on the right, then use arrows to indicate the ordered pairs. Let's call the relation /. / Figure 5.5: A mapping diagram for the relation / depicted in Figure 5.4. Answer: Yes, the relation is a function. In Figure 5.5, note how the domain object 3 is "sent to" or paired with two range objects, 2 and 3. Hence, the relation / is not a function. □ You Try It! The following relation pairs people with their age. Determine if the relation is a function. S"={(Mary,23),(Joe, 18), (Alfonzo,20),(Zoe, 18)}, (Maria, 22), (Chris, 23)} EXAMPLE 5. The following relation pairs automobiles with their gas mileage. Determine if the relation is a function. T = {(Bentley Mulsanne, 18), (Kia Soul, 30), (Lamborghini Gallardo, 20), (Smart Fortwo,41), (Jaguar XF, 23)} Solution: In Figure 5.6, we create a mapping diagram indicating the relation between cars and their gas mileage. Note that each domain object on the left is paired with exactly one range object on the right. Hence, this relation is a function. Bentley Mulsanne Kia Soul Lamborghini Gallardo Smart Fortwo Jaguar XF -> 18 -> 30 -> 20 -s-41 -> 23 Answer: Yes, the relation is a function. Figure 5.6: A mapping diagram for the relation T . □ 5.1. FUNCTIONS 299 You Try It! EXAMPLE 6. The following relation pairs a particular bird with the state that has adopted that bird as its state bird. Determine if the relation is a function. R = { (Yellowhammer, Alabama), (Robin, Connecticut), (Nene, Hawaii), (Robin, Michigan)} Solution: In Figure 5.7, we create a mapping diagram indicating the relation between birds and their state adoptions. Note that the domain object "Robin" is paired with two range objects, "Connecticut" and "Michigan," hence this relation is not a function. R Yellowhammer Robin Nene -> Alabama > Connecticut > Hawaii Michigan The following relation pairs people with the types of cars they own. Determine if the relation is a function. S ={ (Bernard, station wagon), (Tina, truck), (Gilberto, sedan), (Kate, sport utility), (Bernard, sedan), (Kate, minivan)} Figure 5.7: A mapping diagram for the relation R. Answer: No, the relation is not a function. □ Mapping Diagram Notation The goal of this section is to introduce function notation. Let's begin with the mapping diagram in Figure 5.8. / ->2 ->6 Figure 5.8: Mapping diagram . The mapping diagram in Figure 5.8 reveals the following facts: • / maps 1 to 2 or / : 1 — > 2. • / maps 2 to 4 or / : 2 -> 4. 300 CHAPTER 5. POLYNOMIAL FUNCTIONS • f maps 3 to 6 or / : 3 — > 6. • / maps 4 to 8 or / : 4 -» 8. Note how the notation / : 4 — > 8 correlates nicely with the mapping diagram in Figure 5.8. The notation / : 4 — > 8 is read "/ maps 4 to 8" or "/ sends 4 to 8." A closer look at the mapping diagram in Figure 5.8 reveals an interesting pattern. The "rule" seems to be that the relation / doubles each entry in its domain: twice 1 is 2, twice 2 is 4, twice 3 is 6, etc. It's possible to give a general description of this "rule" by writing: / : x — > 2x (5.1) That is, / sends x to twice x, or equivalently, 2x. For example, we might ask "where does / send 15?" To answer this question, we would replace x with 15 in the rule (5.1) to get /: 15 ->2(15), or equivalently, /: 15 -» 30. We could also ask "where does / send — 7?" To answer this question, we would replace x with —7 in the rule (5.1) to get or equivalently, /:-7->2(-7), /:-7->-14. You Try It! Given the rule / : x — > 3x — 5, answer the question "where does / send —2?" Answer: / : — 2 11 EXAMPLE 7. Given the rule / : x — > 2x + 3, answer the question "where does / send 8?" Solution: To find where "/ sends 8," substitute 8 for x in the rule / : x — y 2x + 3 to get /:8^2(8)+3, or equivalently, /:8->19. □ You Try It! Given the rule / : x -» 2x 2 + 5x, answer the question "where does / send 3?" EXAMPLE 8. Given the rule / : x — > x/(x + 3), answer the question "where does / send —1?" 5.1. FUNCTIONS 301 Solution: To find where "/ sends — 1," substitute —1 for x in the rule / : x — > x/(x + 3) to get /:-!■ "' -1 + 3 or equivalently, 1 2 Answer: / : 3 — > 33 D In Examples 7 and 8, note that each time you substitute a value for x in the given rule, you get a unique answer. This means that each object in the domain of / is sent to a unique object in the range of /, making the rules in Examples 7 and 8 functions. This leads us to an itemized description of a function. Rule of Three. A function consists of three parts: • a set of objects which mathematicians call the domain, • a second set of objects which mathematicians call the range. • and a rule that describes how to assign each object in the domain to exactly one object in the range. Function Notation Although the mapping diagram notation / : x -» 3 - 4x is quite easy to understand, the standard function notation used is f(x) = 3 - Ax. With mapping diagram notation, if we want to answer the question "where does / send 12?" , we write: / : x -> 3 - Ax / 12 -> 3-4(12) / 12 -> 3 - 48 / 12 -» -45 Hence, / : 12 — > —45; i.e., / sends 12 to —45. Function notation uses exactly the same concept; i.e, substitute 12 for x. 302 CHAPTER 5. POLYNOMIAL FUNCTIONS Tips for Using Function Notation. 1 . Replace all occurrences of variables in the notation with open parentheses. Leave room between the parentheses to substitute the given value of the variable. 2. Substitute the given values of variables in the open parentheses prepared in the first step. 3. Evaluate the resulting expression according to the Rules Guiding Order of Operations. Given f(x) = 3 — 4x, to evaluate /(12), first restate the function notation, then replace each occurrence of the variable with open parentheses. f(x) = 3-Ax /( ) = 3-4( ) Original function notation. Replace each occurrence of x with open parentheses. Now substitute 12 for x in the open parentheses prepared in the last step. You Try It! 4-5*<-3)-<-3) A 2 10 Figure 5.9: Calculator check. /(12) = 3-4(12) /(12) = 3-48 /(12) = -45 Hence, /(12) = -45; i.e., / sends 12 to -45 Substitute 12 for x in the open parentheses positions. Multiply. Subtract. EXAMPLE 9. Given f{x) =A-5x-x 2 , evaluate /(-3). Solution: Start by replacing each occurrence of the variable x with open parentheses. /( ) bx — x 5( )-( ) 2 Original function notation. Replace each occurrence of x with open parentheses. Now substitute —3 for x in the open parentheses prepared in the last step. /(-3)=4-5(-3)-(-3) 2 /(-3) /(-3) /(-3) 4-5(-3)-9 4+15-9 10 Substitute —3 for x in the open parentheses positions. Evaluate exponent: (—3) Multiply: -5(-3) = 15. Simplify. 9. Thus, /(— 3) = 10. Check this on your calculator (see Figure 5.9). □ 5.1. FUNCTIONS 303 The next example demonstrates one of the advantages of function notation. For example, it is easy to refer to the function in which you want to substitute the given x- value. You Try It! x and g(x) = x 2 — 9, find /(— 1) and Given f(x) = 3x 2 — 20 and EXAMPLE 10. Given f(x) = ff(-2). Solution: We're given two function definitions, / and g, but we're first asked to find /(— 1). This means that we must replace each occurrence of x with —1 in the function f(x) = 5 — x. m /( ) X ( ) Original function notation. Replace each occurrence of x with open parentheses. Now substitute —1 for x in the open parentheses prepared in the last step. /(-l) = 5-(-l) = 5 + 1 = 6 Substitute — 1 for x in the open parentheses position. Add the opposite. Simplify. Thus, /(— 1) = 6. We're next asked to find g{— 2). This means that we must replace each occurrence of x with —2 in the function g(x) = x 2 — 9. g(x) = x 2 - 9( ) = ( ) 2 9 Original function notation. Replace each occurrence of x with open parentheses. Now substitute —2 for x in the open parentheses prepared in the last step. g(x) = 4x + 6/x, find /(-3) and g{2). 5 (-2) = (-2) 2 -9 Substitute —2 for x in the open parentheses position. = 4-9 Exponent first: (-2) 2 = 4. = -5 Simplify. Thus, g{-2) = -5. Answer: /(-3) = 7 and g{2) = = 10 □ Interchanging y and f{x) In most cases, y and f(x) are completely interchangeable. For example, com- pare and contrast the following two examples. 304 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! Sketch the graph of f{x) = -\x-2. Question: Given y y when x equals 5. 3a; + 7, find Solution: Replace x with 5. y = 3x + 7 y = 3(5) + 7 y = 15 + 7 V = 22 Question: Given f(x) = 3x + 7, evaluate /(5). Solution: Replace x with 5. f(x) = 3ai + 7 /(5) = 3(5) + 7 /(5) = 15 + 7 /(5) = 22 In each case, the answer is 22. However, in the first case, the answer y = 22 disguises the fact that an a;- value of 5 was used to arrive at the result. On the other hand, when we use function notation, the final answer /(5) = 22 indicates that we used an a;- value of 5 to determine that the y- value is 22. This is another advantage of function notation. Let's look at one final application demonstrating that y and f(x) are inter- changeable. EXAMPLE 11. Sketch the graph of f(x) = 2x - 3. Solution: Because y and f(x) are interchangeable, the instruction is identical to "sketch the graph of y = 2x — 3." The graph is a line, with slope 2 and y-intercept at (0, —3). Plot the y-intercept at (0, —3), then move up 2 and right 1 to create a line with slope 2 (see Figure 5.10). Note how we've labeled the graph with its equation using function notation. Answer: 5 5 5 [ [ '• - -Z ) ■~> 5 3 \ s f{x) = --x-2 K ' i m / / / / 5 1 i (0 — it ) ^ ■ 2a: -3 Figure 5.10: The graph of f(x) = 2x — 3 is a line. □ 5.1. FUNCTIONS 305 **.!*.**. Exercises •** * ■** In Exercises 1-6, state the domain and range of the given relation. 1. i? = {(7,4), (2,4), (4,2), (8,5)} 2. 5 = {(6,4), (3,3), (2,5), (8,7)} 3. T = {(7,2), (3,1), (9,4), (8,1)} 4.7? = {(0,1), (8,2), (6,8), (9,3)} 5. T = {(4,7), (4,8), (5,0), (0,7)} 6. T = {(9,0), (3,6), (8,0), (3,8)} In Exercises 7-10, state the domain and range of the given relation. 9. I / A B 5 ! C D 5 r, 1 / B A 5 ! c 5 Z3 c 10. B c B D D In Exercises 11-18, determine whether the given relation is a function. 11. R = {(-6, -4), (-4, -4), (1,-4)} 13. T = {(-1,-7), (2,-5), (4,-2)} 12. T = {(-8, -3), (-4, -3), (2, -3)} 14. S = {(-6, -6), (-4, 0), (9, 1)} 306 CHAPTER 5. POLYNOMIAL FUNCTIONS 15. T= {(-9,1), (1,6), (1,8)} 16. S = {(-7,0), (1,1), (1,2)} 17. R = {(-7,-8), (-7,-6), (-5,0)} 18. T = {(-8,-9), (-8, -4), (-5,9)} In Exercises 19-22, determine whether the given relation is a function. 19. 21. .'/ . . C B A 1 L 1- 5 I 20. 1 / A B C 1 L 1- I ' .'/ 22. C B A 1 L 1- 5 I / C B A 1 L 1- ■ 5 I 23. Given f(x) 24. Given f(x) 25. Given f(x) 26. Given f{x) 27. Given f{x) 28. Given f{x) 29. Given f(x) 30. Given f(x) |6x — 9|, evaluate /(8). : |8x — 3|, evaluate /(5). : -2x 2 + 8, evaluate /(3). 3x 2 + x + 6, evaluate /(— 3). -3ie 2 +4ie+1, evaluate /(2). -3x 2 +4x-2, evaluate /(2). ■ |5a; + 9|, evaluate /(-8). : \9x — 6|, evaluate /(4). \Jx — 6, evaluate /(42). Vx + 8, evaluate /(41) 31. Given f(x 32. Given f{x 33. Given f(x) = y/x-7, evaluate /(88) 34. Given f(x 35. Given f(x 36. Given f(x 37. Given f(x 38. Given f{x y/x + 9, evaluate /(16). — 4x + 6, evaluate /(8). — 9x + 2, evaluate /(— 6). — 6x + 7, evaluate /(8). —6x — 2, evaluate /(5). 5.1. FUNCTIONS 307 39. Given f(x) = -2x 2 + 3x + 2 and g(x) = 3x 2 + 5x — 5, evaluate /(3) and g(3). 40. Given f(x) = 3a; 2 — 3x — 5 and g{x) = 2x 2 — 5x — 8, evaluate /(— 2) and g(— 2). 41. Given /(a;) = 6a; — 2 and g(a;) = —8a; + 9, evaluate /(— 7) and g(—7). 42. Given /(a;) = 5a; — 3 and g(a:) = 9a; — 9, evaluate /(— 2) and g(— 2). 43. Given /(x) = 4a; - 3 and g(a;) = -3a; + 8, evaluate /(— 3) and c/(— 3). 44. Given f(x) = 8a; + 7 and g(x) = 2x - 7, evaluate /(— 9) and g(— 9). 45. Given /(a;) = —2a; 2 + 5a; — 9 and g(x) = -2x 2 + 3a; - 4, evaluate /(-2) and g{-2). 46. Given f(x) = -3a; 2 + 5a; - 2 and g(x) = 3a; 2 — 4a; + 2, evaluate /(— 1) and g{— 1). ja- j*- &»■ Answers ••* *•* •** 1. Domain = {2,4,7,8} and Range = {2,4,5} 3. Domain = {3, 7, 8, 9} and Range = {1,2,4} 5. Domain = {0, 4, 5} and Range = {0, 7, 8} 7. Domain = { — 2,2} and Range = {-2,2,4} 9. Domain = {—4,-1,1,2} and Range = {-2,2,4} 11. Function 13. Function 15. Not a function 17. Not a function 19. Function 21. Not a function 23. 39 25. -10 27. -3 29. 31 31. 6 33. 9 35. -26 37. -41 39. /(3) = -7and ff (3) =37 41. /(-7) = -44 and g(-7) = 65 43. /(-3) = -15 and g(-3) = 17 45. /(-2) = -27 and g{-2) = -18 308 CHAPTER 5. POLYNOMIAL FUNCTIONS 5.2 Polynomials We begin with the definition of a term. Term. A term is either a single number (called a constant term) or the product of a number and one or more variables. For example, each of the following is a term. -5 -3a; 2 12y 2 z 3 Ua 2 bc 3 Note how the first term is a single number, while the remaining terms are products of a number and one or more variables. For example, — 3x 2 is the product of —3, x, and x. Coefficient. When a term is a product of a number and one or more variables, the number is called the coefficient of the term. In the case of a term that is a single number, the number itself is called the coefficient. Thus, for example, the coefficients of the terms -5 -3x 2 Yly 2 z 3 13a 2 6c 3 are —5, —3, 12, and 13, respectively. Degree. The degree of a term is the sum of the exponents on each variable of the term. A constant term (single number with no variables) has degree zero. Thus, for example, the degrees of the terms -5 -3x 2 Yly 2 z 3 Ua 2 bc 3 are 0, 2, 5, and 6, respectively. In the last example, note that 13a 2 6c 3 is equivalent to 13a 2 6 1 c 3 , so adding exponents, we get: Degree of 13a 2 6c 3 = Degree of I3a 2 b 1 c 3 =2+1+3 = 6 Monomial. The words monomial and term are equivalent. Thus, -5 -3x 2 Yly 2 z 3 I3a 2 bc 3 are monomials. 5.2. POLYNOMIALS 309 Binomial. A binomial is a mathematical expression containing exactly two terms, separated by plus or minus signs. For example, each of the mathematical expressions 2x + 3y — 3a — 36 xy + 7 — 3x y + 5xy is a binomial. Each expression has exactly two terms. Trinomial. A trinomial is a mathematical expression containing exactly three terms, separated by plus or minus signs. For example, each of the mathematical expressions 2x 2 + 3a; + 7 a 2 + 2ab + b 2 x A - 2x 2 y 2 + 3y 4 is a trinomial. Each expression has exactly three terms. A bicycle has two wheels, a binomial has two terms. A tricycle has three wheels, a trinomial has three terms. But once we get past three terms, the assignment of special names ceases and we use the generic word polynomial, which means "many terms." Polynomial. A polynomial is a many-termed mathematical expression, with terms separated by plus or minus signs. The coefficients of a polynomial are the coefficients of its terms. Each of the previous expressions, 12y 2 z 3 - 3a 2 - 3b 2 x 4 - 2x 2 y 2 + 3y A though assigned the particular names monomial, binomial, and trinomial, re- spectively, are also "many-termed" expressions and can also be called polyno- mials. However, because the word polynomial means "many terms," we can also use the word polynomial to describe mathematical expressions with more than three terms, such as: x 4 - 4x 3 y + 6x 2 y 2 - 4xy 3 + y A The coefficients of x A — 4x 3 y + 6x 2 y 2 — Axy 3 + y A are 1, —4, 6, —4, and 1. Ascending and Descending Powers When asked to simplify a polynomial expression, we should combine any like terms we find, and when possible, arrange the answer in ascending or descend- ing powers. 310 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! Simplify the following polynomial, and arrange your answer in ascending powers of x: 3x 2 - 5a; 3 + 8x + 9a; 2 - 7x + 2x 3 EXAMPLE 1. Simplify the following polynomial expression, arranging your answer in descending powers of x. Once you've completed that task, make a second arrangement, ordering your terms in ascending powers of x. 2x 3 + 7x- 3x 2 + 11a; + 8a; 2 + 11 + 15a; Solution: In order to arrange our answer in descending powers of x, we want to place the term with the highest power of x first and the term with the lowest power of x last. We use the commutative and associative properties to change the order and regroup, then we combine like terms. 2a; 3 + 7x - 3a; 2 + 11a; + 8a; 2 11 + 15a; = 2a; 3 + (-3a; 2 + 8a; 2 ) + (7x + 11a: + 15a;) + 11 = 2a; 3 + 5a; 2 + 33a; +11 Note how the powers of x start at 3, then go down in order. To arrange our final answer in ascending powers of x, we put the lowest power of x first, then the highest power of x last, regrouping and combining like terms. Answer: x + 12a; 2 33^ 2a; 3 + 7x - 3x 2 + 11a; + 8a; 2 + 11 + 15a; = 11 + (7a; + 11a; + 15a;) + (-3a; 2 + 8a; 2 ) + 2a; 3 = ll + 33x + 5x 2 + 2a; 3 Note how we start with the constant term, then the powers of x increase in order. □ When we have a polynomial in a single variable, such as the polynomial in Example 1, arranging the terms in ascending or descending order is fairly straightforward. However, a polynomial in two or more variables is a bit more difficult, and sometimes impossible, to arrange in a decent order. You Try It! Simplify the following polynomial, and arrange your answer in descending powers of x: -ix 2 y 2 + 3xy 3 + 6x 3 y — xy 3 + 2a; 2 y 2 EXAMPLE 2. Simplify the following polynomial expression, then arrange your answer in descending powers of x. 2xy 2 - 6x 2 y + y 3 3xy 2 + 4x 2 y Solution: We'll again use the commutative and associative properties to change the order and regroup, putting the terms with the highest powers of x 5.2. POLYNOMIALS 311 first, then follow with terms containing lower powers of x in order. x 3 + 2xy 2 - 6x 2 y + y 3 - 3xy 2 + 4x 2 y = x 3 + {-6x 2 y + 4x 2 y) + {2xy 2 - 3xy 2 ) + y 3 x 3 - 2x 2 y xy^ + y J Note that this is a very natural order, the powers of x decrease while simulta- neously the powers of y increase. Answer: 6x 3 y - 2x 2 y 2 + 2xy 3 Not all examples will have nice ordering presented in Example 2, with the powers of one variable descending while the powers of the other variable simul- taneously ascends. Sometimes we have to make some very subjective choices on the ordering of terms. □ EXAMPLE 3. Simplify the following polynomial expression, then arrange your answer in some sort of reasonable order. a 3 b 3 + 2a 2 b - 3a 2 b 3 + Aa 3 b 3 + 5a 4 + 3a A b + 6° Solution: Let's try to arrange the terms so that the powers of a descend. Again, we use the commutative and associative properties to change the order and regroup. You Try It! Simplify the following polynomial, and arrange your answer in ascending powers of b: 5a 3 b 2 + 4ab 3 - 2a 2 b+ 3a 3 b 2 - ab 3 3^3 (!%■ 2a 2 b- 3a 2 b 3 Aa 3 b 3 + 5a 4 + 3a 2 b + b 5 = 5a 4 + (a 3 b 3 + 4a 3 b 3 ) + (2a 2 b + 3a 2 b) - 3a 2 b 3 + b b = 5a 4 + 5a 3 6 3 + 5a 2 b - 3a 2 b 3 + b 5 Note that in our final arrangement, the powers of a descend, but the powers of b bounce up and down, but at least we have the powers of a descending. That should help us spot if we've missed a term while simplifying the given problem. Answer: -2a 2 b + 8a 3 b 2 3ab 3 □ The Degree of a Polynomial To find the degree of a polynomial, locate the term of the polynomial having the highest degree. The degree of a polynomial. The the term having the highest degree. of a polynomial is the degree of 312 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! What is the degree of the polynomial 2a; 3 + 8x 2 + 3a; 4 + 2x + 10? Answer: 4 You Try It! What is the degree of the polynomial x 2 y 4 — 6x 2 y 2 + 5x 2 y 5 — 2xy? Answer: 7 Finding the degree of a polynomial of a single variable is pretty easy. EXAMPLE 4. What is the degree of the polynomial a: 3 — 4a; 2 + 5 — 6x + 2x 7 ? Solution: First, let's arrange the polynomial in descending powers of x. 2x' 4cT - 6x + 5 Arranging the polynomial in descending powers of x makes it easier to see that the term of the polynomial with the highest degree is 2a; 7 . Therefore, the degree of the polynomial is 7. □ Finding the degree of a polynomial of more than one variable is a little bit trickier. EXAMPLE 5. What is the degree of the polynomial x — 2x y 3„,7 y 5? Solution: Note that the polynomial is already arranged in descending powers of a;, an arrangement that is probably as good as we are going to get. In the following table, we list the degree of each term. Remember, the degree of any term is found by summing the exponents on its variables. Term Degree x 4 4 -2x i y 7 10 y 5 5 Hence, the term with the highest degree is —2x 3 y 7 , making 10 the degree of the polynomial. □ Polynomial Functions First we define what we mean by a polynomial function. Polynomial function. A polynomial function is a function defined by a rule that assigns to each domain object a range object defined by a polynomial expression. 5.2. POLYNOMIALS 313 Advanced courses, such as multivariate calculus, frequently use polynomial functions of more than one variable such as f(x, y) = x 2 + y 2 . However, in this course, our focus will be on polynomial functions of a single variable, such as p(x) = 3 - ix - 9x 2 and q(x) = x 3 - 9x 2 + 11. You Try It! EXAMPLE 6. Given the polynomial function p(x) p(-3). Solution: To evaluate p(— 3), first restate the function definition, then replace each occurrence of the variable x with open parentheses. 8a; — 11, evaluate Given the polynomial function p(x) = -3x 2 + 7x + A, evaluate p(2). p(x) = x — 8x p( ) = ( ) 3 " 11 Original function definition. ( ) — 1 1 Replace each occurrence of x with open parentheses. Next, substitute —3 for x in the open parentheses prepared in the last step. p(-3) = (-3) s p(-3) = p(-3) = p(-3) = Hence, p(— 3) Figure 5.11). -27-8(-3)- 11 -27+24- 11 -14 11 Substitute —3 for x in the open parentheses positions. Exponent first: (-3) 3 = -27. Multiply: -8(-3) = 24. Add. ( -3)^3-8* <-3)-ll -14 Figure 5.11: check. Calculator -14. You can easily check this result on your calculator (see Answer: 6 The Graph of a Polynomial Function One of the most important polynomial functions in all of mathematics and science is the polynomial having degree two. Quadratic polynomial. The second degree polynomial having the form p(x) = ax 2 + bx + c is called a quadratic polynomial. The graph of this polynomial is called a parabola. □ The parabola is approximately U-shaped. Some open upwards, some open downwards, depending on the sign of the leading term. In Figure 5.12, the leading term of the parabola p(x) = 2x 2 — 8x + 6 has positive two as its 314 CHAPTER 5. POLYNOMIAL FUNCTIONS -5-* Figure 5.12: The graph of p(x) 2x 2 — 8x + 6 opens up. H 5 "' zx «_ r\ sir T V-4- t J t =fc=fe= Figure 5.13: The graph of p(x) —2x 2 — 8a; — 6 opens down. coefficient, so it opens upward. In Figure 5.13, the leading term of the parabola p(x) = — 2x 2 — 8x — 6 has negative two as its coefficient, so it opens downward. Thes tgn of the leading term of p(x) = ax 2 + bx + c determines whether the parabola opens up or down. • If a > 0, the parabola opens upward. • If a < 0, the parabola opens downward. The turning point of a parabola has a special name. The vertex of a parabola. The graph of the second degree polynomial p(x) = ax 2 + bx + c has a single turning point, called the vertex of the parabola. You Try It! Use your graphing calculator to sketch the graph of the quadratic polynomial p(x) = 2x 2 - 5x - 4. EXAMPLE 7. Use your graphing calculator to sketch the graph of the quadratic polynomial p(x) = —3a; 2 + 12a; + 25. Solution: The degree of the polynomial p(x) = —3x 2 + 12a; + 25 is two, so it is a quadratic polynomial and its graph is a parabola. Moreover, its leading term has negative three as its coefficient, so we know that the parabola opens downward. Enter y = -3a; 2 + 12a; + 25 as Yl=-3*XA2+12*X+25 in the Y= menu (see the first image in Figure 5.14), then select 6:ZStandard from the ZOOM menu to produce the third image in Figure 5.14. Note that the graph in Figure 5.14 appears to have the U-shape of a parabola that opens downwards. Its vertex (turning point) is not visible, but one would 5.2. POLYNOMIALS 315 Plotl PlotE Plots WiB-3*X A 2+12*X+ 25 sVs = MEMORV ox Zoom In Zoom Out ZDecinal ZS=iuare ZStandard 7-i-ZTrig Figure 5.14: Sketching the graph of p(x) = —3x 2 + 12a; + 25. surmise that it lies off the top of the screen. We need to adjust the WINDOW parameters so that the vertex of the parabola is visible in the viewing screen. After some experimentation, we settle on the parameters shown in the first image in Figure 5.15, then push the GRAPH button to produce the second image in Figure 5.15. WINDOW Xnin=-10 Xmax=10 Xscl=l Ymin=-50 Vmax=50 Yscl=10 4rXres=i Figure 5.15: Adjust the WINDOW parameters so that the vertex is visible in the viewing screen. In reporting your result on your homework, follow the Calculator Submis- sion Guidelines from Chapter 3, Section2. !J 1. Draw axes with a ruler. 2. Label the horizontal axis x and the vertical axis y. 3. Indicate the WINDOW parameters Xmin, Xmax, Ymin, and Ymax at the end of each axis. 4. Freehand the curve and label it with its equation. 50 p{x) = -3x 2 10 '-50" 10 12a: + 25 Answer: □ 316 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! Use your graphing calculator to sketch the graph of the quadratic polynomial p(x) = x 3 - 14a; 2 + 20a; + 60. Set your window parameters as follows: Xmin = —10, Xmax = 20, Xscl = 1, Ymin = -200, Ymax = 200, and Yscl = 20. Answer ■ ■ / ^ / /I : V When the degree of the polynomial is larger than two, the number of turning points of the graph might increase. This makes for some very interesting curves. In more advanced courses, such as intermediate and college algebra, you will be introduced to a variety of techniques that will help you determine appropriate viewing windows for the graphs of these higher degree polynomials. However, in this introductory section, we will assist you by suggesting a good viewing window for each polynomial, one that will allow you to see all of the turning points of the graph of the polynomial. EXAMPLE 8. Use your graphing calculator to sketch the graph of the polynomial function p(x) = x A — 37a; 2 + 24a; + 180. Set your window parameters as follows: Xmin = —10, Xmax = 10, Xscl = 1, Ymin = —1000, Ymax = 1000, and Yscl = 100. Solution: Enter the polynomial function in Yl of the Y= menu, then enter the suggested window parameters in the WINDOW menu (see Figure 5.16). Plotl PlotE Plots WINDOW WiBX-M-37*>T-2+2 Xnin=-10 4*X+180 Xnax=l@ Wi=l Xscl=l ^Vs= Ymin=-1000 Wh = Vnax=1000 ^Ve= Yscl=100 Ws = l¥.res=W Figure 5.16: Enter the polynomial and adjust the WINDOW parameters. Push the GRAPH button on the top row of your calculator to produce the graph of the polynomial function shown in Figure 5.17. Figure 5.17: The graph of p(x) = x 4 - 37a; 2 + 24a; + 180. Sweet-looking curve! □ 5.2. POLYNOMIALS 317 **.**.**. Exercises -*s -« ■*? In Exercises 1-6, state the coefficient and the degree of each of the following terms. 1. 3v 5 u 6 4. -5c 3 2. -36 5 ;z 8 5. 2u 7 x i d 5 3. -5v 6 6. to 4 c 5 u 7 In Exercises 7-16, state whether each of the following expressions is a monomial, binomial, or trinomial. 7. -76 9 c 3 12. 8w 4 + buv + 3v 4 8. 76 6 c 2 13. 5s 2 + 9£ 7 9. Au + 7v 14. -8a; 6 - 6y 7 10. -36 + 5c 15. 2u 3 - buv - 4v 4 11. 36 4 - 96c + 9c 2 16. 6y 3 - Ayz + 7z 3 In Exercises 17-20, sort each of the given polynomials in descending powers of x. 17. -2a; 7 - 9a; 13 - 6a; 12 - 7a; 17 19. 8a; 6 + 2a; 15 - 3a; 11 - 2a; 2 18. 2a; 4 - 8a; 19 + 3a; 10 - 4a; 2 20. 2a; 6 - 6a; 7 - 7a; 15 - 9a; 18 In Exercises 21-24, sort each of the given polynomials in ascending powers of x. 21. 7a; 17 + 3a; 4 - 2a; 12 + 8a; 14 23. 2a; 13 + 3a; 18 + 8a; 7 + 5a; 4 22. 6a; 18 - 6a; 4 - 2a; 19 - 7a; 14 24. -6x 18 - 8a; 11 - 9a; 15 + 5a; 12 In Exercises 25-32, simplify the given polynomial, combining like terms, then arranging your answer in descending powers of x. 25. -5a; + 3 - 6x 3 + 5a; 2 - 9x + 3 - 3a; 2 + 6a; 3 29. a; 2 + 9a; - 3 + 7a; 2 - 3a; - 8 26. -2a; 3 + 8a; - a; 2 + 5 + 7 + 6a; 2 + 4a; 3 - 9a; 30. -4a; 2 - 6a; + 3 - 3a; 2 + 3a; - 6 27. 4a; 3 + 6a; 2 - 8a; + 1 + 8a; 3 - 7a; 2 + 5a; - 8 31. 8a; + 7 + 2a; 2 - 8a; - 3a; 3 - a; 2 28. -8a; 3 - 2a: 2 - 7a; - 3 + 7a; 3 - 9a; 2 - 8a; + 9 32. -a; 2 + 8 - 7a; + 8a; - 5a; 2 + 4a; 3 318 CHAPTER 5. POLYNOMIAL FUNCTIONS In Exercises 33-44, simplify the given polynomial, combining like terms, then arranging your answer in a reasonable order, perhaps in descending powers of either variable. Note: Answers may vary, depending on which variable you choose to dictate the order. 33. -8a; 2 - 4xz - 2z 2 - 3a; 2 - 8xz + 2z 2 34. -5a; 2 + 9xz - Az 2 - 6a; 2 - Ixz + 7z 2 35. -6u 3 + Auv 2 - 2v 3 - u 3 + 6u 2 v — buv 2 36. 7a 3 6a 2 bab 2 + 4a 3 6a 2 6b 3 37. -46 2 c - 36c 2 - 5c 3 + 96 3 - 36 2 c + 56c 2 38. 46 3 - 66 2 c + 96c 2 - 96 3 - 86c 2 + 3c 3 39. -8y 2 + 6yz - 7z 2 - 2y 2 - 3yz - 9z 2 40. 8x 2 + xy + 3y 2 - x 2 + 7xy + y 2 41. 76 2 c + 86c 2 - 6c 3 - 46 3 + 96c 2 - 6c 3 42. 7x 3 - 9x 2 y + 3y 3 + 7x 3 + 3xy 2 - 7y 3 43. 9a 2 + ac- 9c 2 - 5a 2 - 2ac + 2c 2 44. 7m 2 + 3uv - 6v 2 - 6u 2 + 7uv + 6v 2 In Exercises 45-50, state the degree of the given polynomial. 45. 3x 15 + 4 + 8x 3 - 8x 19 46. -4a; 6 - 7ir 16 - 5 + 3:r 18 47. 7a; 10 - 3a; 18 + 9a; 4 - 6 48. 3a; 16 49. -2 - x' - 5a; D + x^ 50. x 11 + 7a; 16 + 8 - 7a; 10 8x 5 n7 8 + 7 10 51. Given f(x) = 5x 3 + Ax 2 — 6, evaluate /(-I)- 52. Given f(x) = -3a; 3 + 3a; 2 - 9, evaluate /(-I)- 53. Given f(x) = 5a; 4 — 4a; — 6, evaluate /(— 2). 54. Given f{x) = -2x 4 -4x-9, evaluate /(2). 55. Given f(x) = 3a; + 5a; 3 — 9, evaluate /(-2). 56. Given f(x) = —3a; 4 + 2a; 3 — 6, evaluate /(-I)- 57. Given f(x) = 3a; 4 — 5a; 2 + 8, evaluate /("!)■ 58. Given f(x) = —4a; 4 — 5a; 2 — 3, evaluate /(3). 59. Given f(x) = -2a; 3 +4a;-9, evaluate /(2). 60. Given f(x) = 4a; 3 +3a;+7, evaluate f{-2). In Exercises 61-64, use your graphing calculator to sketch the the given quadratic polynomial. In each case the graph is a parabola, so adjust the WINDOW parameters until the vertex is visible in the viewing window, then follow the Calculator Submission Guidelines when reporting your solution on your homework. 61. p(x) = -2x 2 + 8x + 32 62. p(x) = 2x 2 + &x- 18 63. p(x) = 3a; 2 - 8a; - 35 64. p(x) = -Ax 2 - 9a; + 50 5.2. POLYNOMIALS 319 In Exercises 65-68, use your graphing calculator to sketch the polynomial using the given WINDOW parameters. Follow the Calculator Submission Guidelines when reporting your solution on your home- work. 65. p(x) x Xmin Ymin 66. p{x) = — x 3 Xmin = Ymin = Ax 1 -10 -50 4a; 2 11a; + 30 Xmax = Ymax -10 -10 -150 50 27a: - 90 Xmax= -10 Ymax = 50 67. p(x) x 10a; 3 - Ax 1 + 250a; - 525 Xmin Ymin -10 -1000 Xmax= -10 Ymax = 500 68. p(x) -x 2x 6 + 35a; 2 - 36a; - 180 Xmin Ymin -10 -50 Xmax Ymax 10 50 j*- j*- ?*- Answers •** ■*$ ■** 1. Coefficient = 3, Degree =11 3. Coefficient = — 5, Degree = 6 5. Coefficient = 2, Degree = 16 7. Monomial 9. Binomial 11. Trinomial 13. Binomial 15. Trinomial 17. -7a; 17 - 9a; 13 - 6a; 12 - 2a; 7 19. 2a; 15 - 3a; 11 + 8a; 6 - 2a; 2 21. 3a; 4 - 2a; 12 + 8a; 14 + 7a; 17 23. 5a; 4 + 8a; 7 + 2a; 13 + 3a; 18 25. 2a; 2 - 14a; + 6 27. 12a; 3 - a; 2 - 3a; - 7 29. 8a; 2 + 6a;- 11 31. -3ar 3 +a; 2 + 7 33. -11a; 2 - 12a;z 35. -7u 3 + 6u 2 v - uv 2 - 2v 3 37. 96 3 - 76 2 c + 26c 2 - 5c 3 39. -Wy 2 + 3yz- I6z 2 41. -46 3 + 76 2 c + 176c 2 - 12c 3 43. 4a 2 -ac- 7c 2 45. 19 47. 18 49. 10 51. -7 53. 82 55. -1 57. 6 320 CHAPTER 5. POLYNOMIAL FUNCTIONS 63. p(x) = -2x 2 + 8x + 32 65. p(x) = -3x 2 - 8a; - 35 67. p(x) = x 3 - 4x 2 - Ux + 40 y p(x) = x 4 - 10x 3 - 4x 2 + 250a: - 525 V 1000 5.3. APPLICATIONS OF POLYNOMIALS 321 5.3 Applications of Polynomials In this section we investigate real-world applications of polynomial functions. EXAMPLE 1. The average price of a gallon of gas at the beginning of each month for the period starting in November 2010 and ending in May 2011 are given in the margin. The data is plotted in Figure 5.18 and fitted with the following third degree polynomial, where t is the number of months that have passed since October of 2010. p(t) = -0.0080556i 3 + 0.11881£ 2 - 0.30671* + 3.36 (5.2) Use the graph and then the polynomial to estimate the price of a gallon of gas in California in February 2011. You Try It! p(t) 4.50 . 4.40 4.30 4.20 4.10 4 3.90 i » / 3.80 3.70 3.60 3.50 3.40 3.30 V \ 3.20 3.10 ^ i F — 3 ( ) 1 2 3 4 5 6 7 8 ci N )V D ec L in F( ;b M ar A jr M ay J i in Month Price Nov. 3.14 Dec. 3.21 Jan 3.31 Mar. 3.87 Apr. 4.06 May 4.26 Figure 5.18: Fitting gas price versus month with a cubic polynomial. Solution: Locate February (t = 4) on the horizontal axis. From there, draw a vertical arrow up to the graph, and from that point of intersection, a second horizontal arrow over to the vertical axis (see Figure 5.19). It would appear that the price per gallon in February was approximately $3.51. 322 CHAPTER 5. POLYNOMIAL FUNCTIONS p(t) 4.50 . 4.40 4.30 4.20 4.10 4 3.90 <> / 3.80 3.70 3.60 3.50 3.40 3.30 3.20 3.10 i > — 3 ( ) 1 2 3 4 5 6 7 8 ct N )V D ec Jc in F< sb M ar Apr May Jun Figure 5.19: Approximating price of gas during February. Next, we'll use the fitted third degree polynomial to approximate the price per gallon for the month of February, 2011. Start with the function defined by equation 5.2 and substitute 4 for t. p(t) = -0.0080556t 3 + 0.11881* 2 - 0.30671* + 3.36 p(4) = -0.0080556(4) 3 + 0.11881(4) 2 - 0.30671(4) + 3.36 Use the calculator to evaluate p(4) (see Figure 5.20). Rounding to the nearest -0.0080556*4^3+0 .11331*4^2-0.306 71*4+3.36 3.5185616 Figure 5.20: Evaluating p (4). penny, the price in February was $3.52 per gallon. □ 5.3. APPLICATIONS OF POLYNOMIALS 323 EXAMPLE 2. If a projectile is fired into the air, its height above ground at any time is given by the formula where 1 2 y = yo + v t - -gt y = height above ground at time t, 2/o = initial height above ground at time t = 0, vq = initial velocity at time t = 0, g = acceleration due to gravity, t = time passed since projectile's firing. (5.3) You Try It! If a projectile is launched with an initial velocity of 60 meters per second from a rooftop 12 meters above ground level, at what time will the projectile first reach a height of 150 meters? If a projectile is launched with an initial velocity of 100 meters per second (100 m/s) from a rooftop 8 meters (8 m) above ground level, at what time will the projectile first reach a height of 400 meters (400m)? Note: Near the earth's surface, the acceleration due to gravity is approximately 9.8 meters per second per second (9.8 (m/s)/s or 9.8 m/s 2 ). Solution: We're given the initial height is j/o = 8m vq = 100 m/s, and the acceleration due to gravity is g the initial velocity is 9.8 m/s 2 . Substitute these values in equation 5.3, then simplify to produce the following result: y = yo + v t 1 2 -gt y = 8 + 100i - -(9-8)t 2 y = 8 + WOt - 4.9i 2 Enter y lOOt - AM 2 as Y1=8+100*X-4.9*XA2 in the Y= menu (see the first image in Figure 5.21). After some experimentation, we settled on the WINDOW parameters shown in the second image in Figure 5.21. Push the GRAPH button to produce the graph of y = 8 + lOOt — 4.9£ 2 shown in the third image Figure 5.21. In this example, the horizon- tal axis is actually the t-axis. So when we set Xmin and Xmax, we're actually setting bounds on the t-axis. Plotl PlotE Plots WiB8+100*X-4.9* nVs= WINDOW Xnin=0 Xnax=25 Xscl=5 Ymin=-100 Ymax=600 Yscl=50 4,Xres=l Figure 5.21: Sketching the graph of y = 8 + WOt - 4.9t 2 . 324 CHAPTER 5. POLYNOMIAL FUNCTIONS To find when the projectile reaches a height of 400 meters (400 m), substi- tute 400 for y to obtain: 400 = 8 + 100* - 4.9r (5.4) Enter the left-hand side of Equation 5.4 into Y2 in the Y= menu, as shown in the first image in Figure 5.22. Push the GRAPH button to produce the result shown in the second image in Figure 5.22. Note that there are two points of intersection, which makes sense as the projectile hits 400 meters on the way up and 400 meters on the way down. Plotl PlotE Plots WiB8+100*X-4.9* vViB400 sVs=i Wh = Intersection K=£.292£3£9 Y=400 Figure 5.22: Determining when the object first reaches 400 meters. The parabola shown in Figure 5.23 is not the actual flight path of the projectile. The graph only predicts the height of the projectile as a function of time. To find the first point of intersection, select 5:intersect from the CALC menu. Press ENTER in response to "First curve," then press ENTER again in response to "Second curve." For your guess, use the arrow keys to move the cursor closer to the first point of intersection than the second. At this point, press ENTER in response to "Guess." The result is shown in the third image in Figure 5.22. The projectile first reaches a height of 400 meters at approximately 5.2925359 seconds after launch. -100 5.2925359 Figure 5.23: Reporting your graphical solution on your homework. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. 5.3. APPLICATIONS OF POLYNOMIALS 325 • Label the horizontal and vertical axes with t and y, respectively (see Figure 5.23). Include the units (seconds (s) and meters (m)). • Place your WINDOW parameters at the end of each axis (see Figure 5.23) . Include the units (seconds (s) and meters (m)). • Label each graph with its equation (see Figure 5.23). • Draw a dashed vertical line through the first point of intersection. Shade and label the point (with its t- value) where the dashed vertical line crosses the i-axis. This is the first solution of the equation 400 = 8 + lOOt — AM 2 (see Figure 5.23). Rounding to the nearest tenth of a second, it takes the projectile approximately t ~ 5.3 seconds to first reach a height of 400 meters. The phrase "shade and label the point" means fill in the point on the i-axis, then write the i-value of the point just below the shaded point. Answer: pa 3.0693987 seconds □ Zeros and x-intercepts of a Function Recall that f(x) and y are interchangeable. Therefore, if we are asked to find where a function is equal to zero, then we need to find the points on the graph of the function that have a y- value equal to zero (see Figure 5.24). . , f (-3 0) (" 1,0 ) |(3, 0) r -5 / i / \ / \ / / -5- Figure 5.24: Locating the zeros of a function. Zeros and x-intercepts. The points where the graph of / crosses the x-axis are called the x-intercepts of the graph of /. The x-value of each x-intercept is called a zero of the function /. 326 CHAPTER 5. POLYNOMIAL FUNCTIONS The graph of / crosses the x-axis in Figure 5.24 at (—3,0), (—1,0), and (3, 0). Therefore: • The x-intercepts of / are: (—3,0), (—1,0), and (3,0) • The zeros of / are: —3, —1, and 3 Key idea. A function is zero where its graph crosses the x-axis. You Try It! Find the zero(s) of the function f(x) = 2.6x — 9.62. EXAMPLE 3. Find the zero(s) of the function f(x) = 1.5a; + 5.25. Algebraic solution: Remember, /(x) = 1.5a; + 5.25 and y = 1.5a; + 5.25 are equivalent. We're looking for the value of x that makes y = or f(x) = 0. So, we'll start with fix) = 0, then replace /(x) with 1.5a; + 5.25. We want the value of x that makes the function equal to zero. Replace f{x) with 1.5a; + 5.25. /(*) = 1.5a; + 5.25 = Now we solve for x. 1.5a; = -5.25 x = -5.25 1.5 x = -3.5 Subtract 5.25 from both sides. Divide both sides by 1.5. Divide: -5.25/1.5 = -3.5 Check. Substitute —3.5 for x in the function f(x) = 1.5a; + 5.25. /(x) = 1.5a; + 5.25 /(-3.5) = 1.5(-3.5) + 5.25 /(-3.5) = -5.25 + 5.25 /(-3.5) = The original function. Substitute —3.5 for x. Multiply: 1.5(-3.5) = -5.25. Add. Note that —3.5, when substituted for x, makes the function f(x) = 1.5a; + 5.25 equal to zero. This is why —3.5 is called a zero of the function. Graphing calculator solution. We should be able to find the zero by sketch- ing the graph of / and noting where it crosses the x-axis. Start by loading the function f(x) = 1.5x + 5.25 into Yl in the Y= menu (see the first image in Figure 5.25). Select 6:ZStandard from the ZOOM menu to produce the graph of / (see the second image in Figure 5.25). Press 2ND CALC to open the the CALCULATE menu (see the third image in Figure 5.25). To find the zero of the function /: 5.3. APPLICATIONS OF POLYNOMIALS 327 Plotl PlotE Plots WiB1.5*X+5.25 -■■Vi = gilEfiilEiU? 1 : yalue zero : riininuri 4: maximun 5: inter-sect fcidy/dx 7:^fCx>dx Figure 5.25: Finding the zero of f(x) = 1.5a; + 5.25. 1. Select 2:zero from the CALCULATE menu. The calculator responds by asking for a "Left Bound?" (see the first image in Figure 5.26). Use the left arrow button to move the cursor so that it lies to the left of the IE-intercept of / and press ENTER. 2. The calculator responds by asking for a "Right Bound?" (see the second image in Figure 5.26). Use the right arrow button to move the cursor so that it lies to the right of the x-intercept of / and press ENTER. 3. The calculator responds by asking for a "Guess?" (see the third image in Figure 5.26). As long as your cursor lies between the left- and right-bound marks at the top of the screen (see the third image in Figure 5.26), you have a valid guess. Since the cursor already lies between the left- and right-bounds, simply press ENTER to use the current position of the cursor as your guess. Y1=i.E*K+E.£! Left Bound? K=0 Y=£.2£ Y1=i.E*K+E.£E / RiSht Bound? K=-4.2££319 Y=-i.i32979 Y1=i.E*K+E.£E a / Gut::? K=-2.££319i Y=1.H20212H Figure 5.26: Using 2:zero from the CALCULATE menu. The calculator responds by approximating the zero of the function as shown in Figure 5.27. / StKO K=-3.£ Y=0 Figure 5.27: —3.5 is a zero of /. Note that the approximation found using the calculator agrees nicely with the zero found using the algebraic technique. Answer: 3.7 □ 328 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! If a projectile is launched with an initial velocity of 60 meters per second from a rooftop 12 meters above ground level, at what time will the projectile return to ground level? In this example, the horizontal axis is actually the i-axis. So when we set Xmin and Xmax, we're actually setting bounds on the t-axis. EXAMPLE 4. How long will it take the projectile in Example 2 to return to ground level? Solution: In Example 2, the height of the projectile above the ground as a function of time is given by the equation y = 8 + 100t- AM 2 . When the projectile returns to the ground, its height above ground will be zero meters. To find the time that this happens, substitute y = in the last equation and solve for t. 100t-4.9r Enter the equation y = 8 + 100£ — 4.9£ 2 into Yl in the Y= menu of your calculator (see the first image in Figure 5.28), then set the WINDOW param- eters shown in the second image in Figure 5.28. Push the GRAPH button to produce the graph of the function shown in the third image in Figure 5.28. Plotl PlotE Plots WiB8+100*X-4.9* Wh = WINDOW Xnin=0 Xmax=25 Xscl=5 Ymin=-100 Ymax=600 Yscl=50 l¥.res=W Figure 5.28: Sketching the graph of y lOOi - 4.9^ To find the time when the projectile returns to ground level, we need to find where the graph of y = 8 + lOOt — A.9t 2 crosses the horizontal axis (in this case the i-axis) . Select 2:zero from the CALC menu. Use the arrow keys to move the cursor slightly to the left of the ^-intercept, then press ENTER in response to "Left bound." Move your cursor slightly to the right of the ^-intercept, then press ENTER in response to "Right bound." Leave your cursor where it is and press ENTER in response to "Guess." The result is shown in Figure 5.29. Figure 5.29: Finding the time when the projectile hits the ground. 5.3. APPLICATIONS OF POLYNOMIALS 329 • Label the horizontal and vertical axes with t and y, respectively (see Figure 5.30). Include the units (seconds (s) and meters (m)). • Place your WINDOW parameters at the end of each axis (see Figure 5.30). • Label the graph with its equation (see Figure 5.30). • Draw a dashed vertical line through the ^-intercept. Shade and label the i-value of the point where the dashed vertical line crosses the i-axis. This is the solution of the equation = 8 + lOOi — 4.9f 2 (see Figure 5.30). y(m) 600 1 -100 20.487852 Figure 5.30: Reporting your graphical solution on your homework. Rounding to the nearest tenth of a second, it takes the projectile approximately t rs 20.5 seconds to hit the ground. Answer: « 12.441734 seconds □ 330 CHAPTER 5. POLYNOMIAL FUNCTIONS ;».;-». t± Exercises •*;■*; -*i 1. A firm collects data on the amount it spends on advertising and the resulting revenue collected by the firm. Both pieces of data are in thousands of dollars. x (advertising costs) 5 15 20 25 30 R (revenue) 6347 6524 7591 8251 7623 7478 The data is plotted then fitted with the following second degree polynomial, where x is the amount invested in thousands of dollars and R(x) is the amount of revenue earned by the firm (also in thousands of dollars). R(x) -4.1a;^ + 166.8a; + 6196 Use the graph and then the polynomial to estimate the firm's revenue when the firm invested $10,000 in advertising. R (thousands of dollars) 8,500 8,000 7,500 7,000 6,500 6,000 (i .A it > x (thousands of dollars) 5 10 15 20 25 30 2. The table below lists the estimated number of aids cases in the United States for the years 1999- 2003. Year AIDS Cases 1999 2000 2002 2003 41,356 41,267 41,289 43,171 The data is plotted then fitted with the following second degree polynomial, where t is the number of years that have passed since 1998 and N(t) is the number of aids case reported t years after 1998. N(t) = 345. 14i 2 - 1705.7* + 42904 Use the graph and then the polynomial to estimate the number of AIDS cases in the year 2001. 5.3. APPLICATIONS OF POLYNOMIALS 331 N (Number of AID cases) 45,000 44,000 43,000 42,000 41,000 40,000 1 t (Years since 1998) 3. The following table records the concentration (in milligrams per liter) of medication in a patient's blood after indicated times have passed. Time (Hours) 0.5 1 1.5 2.5 Concentration (mg/L) 78.1 99.8 84.4 15.6 The data is plotted then fitted with the following second degree polynomial, where t is the number of hours that have passed since taking the medication and C(t) is the concentration (in milligrams per liter) of the medication in the patient's blood after t hours have passed. C(t) -56.214r + 139.3K + 9.35 Use the graph and then the polynomial to estimate the the concentration of medication in the patient's blood 2 hours after taking the medication. C (mg/L) , i 1 \ oU " ( ( / DU " 4U " zu ■ \« t n i 1 1 \ -> t (Hours) 332 CHAPTER 5. POLYNOMIAL FUNCTIONS 4. The following table records the population (in millions of people) of the United States for the given year. Year 1900 1920 1940 1960 1980 2000 2010 Population (millions) 76.2 106.0 132.2 179.3 226.5 281.4 307.7 The data is plotted then fitted with the following second degree polynomial, where t is the number of years that have passed since 1990 and P(t) is the population (in millions) t years after 1990. P(t) = 0.008597* 2 + 1, 1738i + 76.41 Use the graph and then the polynomial to estimate the the population of the United States in the year 1970. P (millions of people) 20 40 60 80 100 120 t (years since 1990) If a projectile is launched with an ini- tial velocity of 457 meters per second (457 m/s) from a rooftop 75 meters (75 m) above ground level, at what time will the projectile first reach a height of 6592 me- ters (6592 m)? Round your answer to the nearest second. Note: The accelera- tion due to gravity near the earth's sur- face is 9.8 meters per second per second (9.8 m/s 2 ). If a projectile is launched with an ini- tial velocity of 236 meters per second (236 m/s) from a rooftop 15 meters (15 m) above ground level, at what time will the projectile first reach a height of 1838 me- ters (1838 m)? Round your answer to the nearest second. Note: The accelera- tion due to gravity near the earth's sur- face is 9.8 meters per second per second (9.8 m/s 2 ). If a projectile is launched with an ini- tial velocity of 229 meters per second (229 m/s) from a rooftop 58 meters (58 m) above ground level, at what time will the projectile first reach a height of 1374 me- ters (1374 m)? Round your answer to 5.3. APPLICATIONS OF POLYNOMIALS 333 the nearest second. Note: The accelera- tion due to gravity near the earth's sur- face is 9.8 meters per second per second (13.8 m/s 2 ). If a projectile is launched with an ini- tial velocity of 234 meters per second (234 m/s) from a rooftop 16 meters (16 m) above ground level, at what time will the projectile first reach a height of 1882 me- ters (1882m)? Round your answer to the nearest second. Note: The accelera- tion due to gravity near the earth's sur- face is 9.8 meters per second per second (9.8 m/s 2 ). In Exercises 9-12, first use an algebraic technique to find the zero of the given function, then use the 2:zero utility on your graphing calculator to locate the zero of the function. Use the Calculator Submission Guidelines when reporting the zero found using your graphing calculator. 9. f{x) = 3.25a; - 4.875 10. f(x) = 3.125 -2.5a; 11. f(x) = 3.9- 1.5a; 12. f(x) = 0.75a; + 2.4 13. If a projectile is launched with an ini- tial velocity of 203 meters per second (203 m/s) from a rooftop 52 meters (52 m) above ground level, at what time will the projectile return to ground level? Round your answer to the nearest tenth of a sec- ond. Note: The acceleration due to grav- ity near the earth's surface is 9.8 meters per second per second (9.8 m/s 2 ). 14. If a projectile is launched with an ini- tial velocity of 484 meters per second (484 m/s) from a rooftop 17 meters (17 m) above ground level, at what time will the projectile return to ground level? Round your answer to the nearest tenth of a sec- ond. Note: The acceleration due to grav- ity near the earth's surface is 9.8 meters per second per second (9.8 m/s 2 ). 15. If a projectile is launched with an ini- tial velocity of 276 meters per second (276 m/s) from a rooftop 52 meters (52 m) above ground level, at what time will the projectile return to ground level? Round your answer to the nearest tenth of a sec- ond. Note: The acceleration due to grav- ity near the earth's surface is 9.8 meters per second per second (9.8 m/s 2 ). 16. If a projectile is launched with an ini- tial velocity of 204 meters per second (204 m/s) from a rooftop 92 meters (92 m) above ground level, at what time will the projectile return to ground level? Round your answer to the nearest tenth of a sec- ond. Note: The acceleration due to grav- ity near the earth's surface is 9.8 meters per second per second (9.8 m/s 2 ). ;*.;-». j». Answers ■** -** •** 1. Aprpoximately $7,452 3. Approximately 63 mg/L 5. 17.6 seconds 7. 6.7 seconds 9. Zero: 1.5 11. Zero: 2.6 13. 41.7 seconds 15. 56.5 seconds 334 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! Simplify: (3s 2 -2si + 4i 2 ) + (s 2 + 7 at - 5t 2 ) Answer: 4s 2 + 5st — t 2 5.4 Adding and Subtracting Polynomials In this section we concentrate on adding and subtracting polynomial expres- sions, based on earlier work combining like terms in Ascending and Descending Powers on page 309. Let's begin with an addition example. EXAMPLE 1. Simplify: (a 2 + 3ab - b 2 ) + (4a 2 + llab - 9b 2 ) Solution: Use the commutative and associative properties to change the order and regroup. Then combine like terms. (a 2 + 3ab - b 2 ) + (4a 2 + llab - 96 2 ) = (a 2 + 4a 2 ) + (3a6 + llab) + (-b 2 - 9b 2 ) = 5a 2 + Uab- 10b 2 a Let's combine some polynomial functions. You Try It! Given f(x) = 2x 2 + 9x - 5 and g(x) 4a; + 3, simplify f(x) + g(x). Answer: x 2 + 5x — 2 EXAMPLE 2. Given f(x) =3x 2 -4x-8 and g(x) = x 2 - lis + 15, simplify f(x)+g(x). Solution: First, replace fix) and g{x) with their definitions. Be sure to surround each polynomial with parentheses, because we are asked to add all of f{x) to all of g(x). f{x) + g(x) = (3a; 2 - 4sc - 8) + (a; 2 - 11a; + 15) Now use the commutative and associative properties to change the order and regroup. Combine like terms. = (3a; 2 + a; 2 ) + (-4a; - 11a;) + (-8 + 15) = 4a; 2 - 15a: + 7 Hence, fix) + gix) = 4a; 2 — 15a; + 7. □ If you are comfortable skipping a step or two, it is not necessary to write down all of the steps shown in Examples 1 and 2. Let's try combining like terms mentally in the next example. 5.4. ADDING AND SUBTRACTING POLYNOMIALS 335 EXAMPLE 3. Simplify: (a; 3 - 2x 2 y + 3xy 2 + y 3 ) + (2x 3 - Ax 2 y - 8xy 2 + by 3 ) You Try It! Simplify: (-ba 2 b + 4ab - 3ab 2 )- {2a 2 b + Tab - ab 2 ) Solution: If we use the associative and commutative property to reorder and regroup, then combine like terms, we get the following result. (x 3 - 2x 2 y + 3xy 2 + y 3 ) + (2x 3 - Ax 2 y - 8xy 2 + 5y 3 ) = {x 3 + 2x 3 ) + {-2x 2 y - 4x 2 y) + (3a;?/ 2 - 8xy 2 ) + (y 3 + by 3 ) = 2.x 3 - 6x 2 y - 5xy 2 + Qy 3 However, if we can combine like terms mentally, eliminating the middle step, it is much more efficient to write: (a; 3 - 2x 2 y + 3xy 2 + y 3 ) + (2x 3 - Ax 2 y - 8xy 2 + by 3 ) = 3x — 6x y — bxy + 6y Answer: -3a 2 b+lla6-4a6 2 d a Negating a Polynomial Before attempting subtraction of polynomials, let's first address how to negate or "take the opposite" of a polynomial. First recall that negating is equivalent to multiplying by — 1 . We can use this property to simplify — (a + b). First, negating is identical to multiplying by —1. Then we can distribute the — 1. — (a + b) = (— l)(a + b) Negating is equivalent to multiplying by-1. = (-l)a + (-1)6 Distribute the -1. = —a + (—b) Simplify: (— l)a = —a and (—1)6 = — b. = —a — b Subtraction means add the opposite. 336 CHAPTER 5. POLYNOMIAL FUNCTIONS Thus, — (a + b) = —a — b. However, it is probably simpler to note that the minus sign in front of the parentheses simply changed the sign of each term inside the parentheses. Negating a sum. When negating a sum of terms, the effect of the minus sign is to change each term in the parentheses to the opposite sign. -(a + < -a — b You Try It! Simplify: -(2a; 2 - 3a; + 9) Let's look at this principle in the next example. EXAMPLE 4. Simplify: -(-3a; 2 + 4a; - 8) Solution: First, negating is equivalent to multiplying by —1. Then distribute the -1. (-3a; 2 + 4a; -8) = (-l)(-3a; 2 + 4a;- Negating is equivalent to multiplying by —1. (-l)(-3a; 2 ) + (-l)(4a;) - (-1)(8) Distribute the -1. 3a; 2 + (-4a;) - (- 3a; 2 - 4x + : Simplify: (-l)(-3a; 2 ) = 3a; 2 (-l)(4a;) = -4a;, and and (-1)(8) = -8. Subtraction means add the opposite. Answer: — 2x 2 + 3a; — 9 Alternate solution: As we saw above, a negative sign in front of a parentheses simply changes the sign of each term inside the parentheses. So it is much more efficient to write -(-3a; 2 +4a;- 3a; 2 - 4a; + I simply changing the sign of each term inside the parentheses. □ Subtracting Polynomials Now that we know how to negate a polynomial (change the sign of each term of the polynomial), we're ready to subtract polynomials. 5.4. ADDING AND SUBTRACTING POLYNOMIALS 337 You Try It! EXAMPLE 5. Simplify: (y 3 - 3y 2 z + Ayz 2 + z 3 ) - (2y 3 - 8y 2 z + 2yz 2 - 8z 3 ) Solution: First, distribute the minus sign, changing the sign of each term of the second polynomial. [y 3 - 3y 2 z + Ayz 2 + z 3 ) - {2y 3 - 8y 2 z + 2yz 2 - 8z 3 ) = y 3 - 3y 2 z + Ayz 2 + z 3 - 2y 3 + 8y 2 z - 2yz 2 + 8z 3 Regroup, combining like terms. You may perform this next step mentally if you wish. = (y 3 - 2y 3 ) + {-3y 2 z + 8y 2 z) + (Ayz 2 - 2yz 2 ) + (z 3 + 8z 3 ) = -y 3 + 5y 2 z + 2yz 2 + 9z 3 Simplify: (Aa 2 b + 2ab-7ab 2 )- (2a 2 b -ab- 5ab 2 ) Answer: 2a 2 b + 3ab - 2ab 2 D Let's subtract two polynomial functions. You Try It! 6ar 7x- 11, EXAMPLE 6. Given p(x) = -5x 3 + 6x - 9 and q(x) simplify p(x) — q(x). Solution: First, replace p(x) and q(x) with their definitions. Because we are asked to subtract all of q(x) from all of p(x), it is critical to surround each polynomial with parentheses. p{x) - q(x) = {-5x 3 + 6x - 9) - (6x 2 - 7x - 11) Distribute the minus sign, changing the sign of each term in the second poly- nomial, then regroup and combine like terms. = -5x 3 + 6x - 9 - 6a; 2 + 7x + 11 = -5x 3 - 6x 2 + (6x + 7x) + (-9 + 11) = -5x 3 -6a; 2 + 13a: + 2 However, after distributing the minus sign, if we can combine like terms men- tally, eliminating the middle step, it is much more efficient to write: p(x) - q(x) = (-5a; 3 + 6x - 9) - (6a; 2 - 7x - 11) = -5a; 3 + 6a; - 9 - 6a; 2 + 7a; + 11 = -5a; 3 - 6a; 2 + 13a; + 2 Given f(x) = 3x 2 + 9x - 4 and g(x) = — 5x 2 + Ax — 6, simplify f(x) - g(x). Answer: 8a; 2 + 5a; + 2 □ 338 CHAPTER 5. POLYNOMIAL FUNCTIONS Some Applications Recall that the area of a rectangle having length L and width W is found using the formula A = LW . The area of a square having side s is found using the formula A = s 2 (see Figure 5.31). W A = LW L Figure 5.31: Area formulae for the rectangle and square. You Try It! Find the area of the square shown below by summing the area of its parts. x 4 EXAMPLE 7. Find the area of the square in Figure 5.32 by summing the area of its parts. Solution: Let's separate each of the four pieces and label each with its area (see Figure 5.33). X 3 3 X Figure 5.32: Find the sum of the parts. A 2 = 3x Figure 5.33: Finding the area of each of the four parts. The two shaded squares in Figure 5.33 have areas A\ = x 2 and A% = 9, respec- tively. The two unshaded rectangles in Figure 5.33 have areas A 2 = 3x and A4 = 3x. Summing these four areas gives us the area of the entire figure. Answer: 8a; +16 3x + 9 + 3x 6x + 9 □ 5.4. ADDING AND SUBTRACTING POLYNOMIALS 339 You Try It! EXAMPLE 8. Ginger runs a business selling wicker baskets. Her business costs for producing and selling x wicker baskets are given by the polynomial function C(x) = 100 + 3a; — 0.02a; 2 . The revenue she earns from selling x wicker baskets is given by the polynomial function R(x) = 2.75a;. Find a formula for P{x), the profit made from selling x wicker baskets. Use your formula to determine Ginger's profit if she sells 123 wicker baskets. Solution: The profit made from selling x wicker baskets is found by subtract- ing the costs incurred from the revenue received. In symbols: P{x) = R(x) - C(x) Next, replace R(x) and C(x) with their definitions. Because we are supposed to subtract all of the cost from the revenue, be sure to surround the cost polynomial with parentheses. P(x) = 2.75a; - (100 + 3a; - 0.02a; 2 ) Distribute the minus sign and combine like terms. = 2.75a;- 100 - 3x + 0.02a; 2 = 0.02a; 2 - 0.25a; - 100 Thus, the profit function is P{x) = 0.02a; 2 - 0.25a; - 100. Next, to determine the profit if 123 wicker baskets are sold, substitute 123 for x in the profit function P{x). P(x) = 0.02a; 2 - 0.25a; - 100 P(123) = 0.02(123) 2 - 0.25(123) - 100 You can now use your graphing calculator to determine the profit (see Figure 5.34). Hence, the profit made from selling 123 wicker baskets is $171.83. The costs for producing and selling x widgets are given by the polynomial function C(x) =50 + 5a; -0.5a; 2 , and the revenue for selling x widgets is given by the polynomial function R{x) = 3.5a;. Determine the profit if 75 widgets are sold. 0.02*123^2-0.25* 123-100 171. S3 Figure 5.34: Determining the profit from selling 123 wicker baskets. Answer: $2,650 □ 340 CHAPTER 5. POLYNOMIAL FUNCTIONS **> t* t* Exercises ■** ■*$ •*$ In Exercises 1-8, simplify the given expression. Arrange your answer in some sort of reasonable order. 1. (-8r 2 t + 7rt 2 + 3t 3 ) + (9r 3 + 2rt 2 + At 3 ) 5. (-2r 2 + 7rs + 4s 2 ) + (-9r 2 + 7rs - 2s 2 ) 2. (-a 3 - 8ac 2 - 7c 3 ) + (-7a 3 - 8a 2 c + 8ac 2 ) 6. (-2r 2 + 3rt - 4i 2 ) + (7r 2 + Art - It 2 ) 3. (7x 2 - 6x - 9) + (8a; 2 + 10sc + 9) 7. (-8y 3 - 3y 2 z - 6z 3 ) + (-3y 3 + 7y 2 z - 9yz 2 ) 4. (-7x 2 + 5x - 6) + (-10s 2 - 1) 8. {7y 2 z + 8yz 2 + 2z 3 ) + (8y 3 - 8y 2 z + 9yz 2 ) In Exercises 9-14, simplify the given expression by distributing the minus sign. 9. -(5x 2 - 4) 12. -(7w 3 - 8u 2 v + 6uv 2 + 5v 3 ) 10. -(-8a; 2 -5) 13. ~{-5x 2 + 9xy + 6y 2 ) 11. -(9r 3 - Ar 2 t - 3rt 2 + At 3 ) 14. -(-4u 2 - Guv + 5v 2 ) In Exercises 15-22, simplify the given expression. Arrange your answer in some sort of reasonable order. 15. (-u 3 - Au 2 w + 7w 3 ) - (u 2 w + uw 2 + 3w 3 ) 19. (-7r 2 - 9rs - 2s 2 ) - (-8r 2 - 7rs + 9s 2 ) 16. (-b 2 c + 86c 2 + 8c 3 ) - (6b 3 + b 2 c - Abe 2 ) 20. (-4a 2 + bab - 2b 2 ) - (-8a 2 + 7a6 + 26 2 ) 17. (2y 3 - 2y 2 2 + 3z 3 ) - (-8y 3 + hyz 2 - 3z 3 ) 21. (10a; 2 + 2x - 6) - (-8a; 2 + 14a; + 17) 18. (4a 3 + 6ac 2 + 5c 3 ) - (2a 3 + 8a 2 c - 7ac 2 ) 22. (-5a; 2 + 19a; - 5) - (-15a; 2 + 19a; + 8) In Exercises 23-28, for the given polynomial functions f(x) and g(x), simplify f(x) + g(x). Arrange your answer in descending powers of x. 23. f(x) = -2x 2 +9x + 7 26. 24. f(x) = -8x 3 +6x-9 27. /(*) = ■■ -2a; 2 + 9a; + 7 g(x) = : 8a; 3 - 7a; 2 + 5 m = : -8x 3 +6a;-9 g(x) = : a; 3 - a; 2 + 3a; /(*) = : 5a; 3 - 5a; 2 + 8a; g(x) = : 7a; 2 - 2a; - 9 25. f(x) = bx° -bx* + 8x 28. m = -a; 2 - f 8a; +1 g(x) = -7a; 3 + 8a;-9 m = -3a; 2 -8a; -9 g(x) = 5a; 2 - - 4a; + 4 /(*) = -3a; 2 + a;-8 g(x) = 7a; 2 - 9 5.4. ADDING AND SUBTRACTING POLYNOMIALS 341 In Exercises 29-34, for the given polynomial functions f(x) and g(x), simplify f(x) — g(x). Arrange your answer in descending powers of x. 29. 30. 31. 9{x) -6x J 7a; + 7 -3a; 3 - 3x z 8x f( x ) = 5a; 3 - 5x + 4 g (x) = -8x 3 - 2x 2 - f(x) = 12a; 2 -5a; + 4 g(x) = 8x 2 - 16a;- 7 3x 32. 33. 34. m = -7x 2 + Ylx + 17 g{x) = -10a; 2 - 17 f{x) = -3a; 3 - 4a; + 2 g(x) = -4a; 3 - 7a; 2 + 6 M = -9a; 2 + 9a; + 3 7^,3 | njl | e In Exercises 35-36, find the area of the given square by summing the areas of its four parts. 35. 36. X 5 5 X X 7 7 X 37. Rachel runs a small business selling wicker baskets. Her business costs for produc- ing and selling x wicker baskets are given by the polynomial function C(x) = 232 + 7a;— 0.0085a; 2 . The revenue she earns from selling x wicker baskets is given by the polynomial function R(x) = 33.45a;. Find a formula for P{x), the profit made from selling x wicker baskets. Use your formula to determine Rachel's profit if she sells 233 wicker baskets. Round your answer to the nearest cent. 38. Eloise runs a small business selling baby cribs. Her business costs for producing and selling x baby cribs are given by the polynomial function C(x) = 122 + 8a; — 0.0055a; 2 . The revenue she earns from sell- ing x baby cribs is given by the polyno- mial function R(x) = 33.45a;. Find a for- mula for P(x), the profit made from sell- ing x baby cribs. Use your formula to de- termine Eloise's profit if she sells 182 baby cribs. Round your answer to the nearest cent. 342 CHAPTER 5. POLYNOMIAL FUNCTIONS £»- j*- &*■ Answers •*$ *** •** 1. 9r 3 - 8r 2 t + 9rt 2 + 7£ 3 21. 18x 2 - 12a; - 23 3. 15a; 2 + 4a; 23. 8a; 3 - 9a; 2 + 9x + 12 5. -llr 2 + 14rs + 2s 2 r ^3 , o„2 7. -lly 3 + 4y 2 z - 9yz 2 - 6z 3 9. -5a; 2 + 4 11. -9r 3 + 4r 2 t + 3r£ 2 - At 3 13. 5a; 2 — 9a;y — 6y 2 15. — u 3 — 5u 2 w — uw 2 + Aw 3 17. 10y 3 - 2y 2 z - 5yz 2 + 6z 3 19. r 2 - 2rs- lis 2 25. 5x 3 + 2aT + 6a; - 9 27. 2a; 2 - 12a; - 5 29. -3a; 3 + 3a; 2 + x + 7 31. 4x 2 + lla;+ll 33. a; 3 + 7a; 2 - 4a; - 4 35. a; 2 + 10a; + 25 37. $6,392.31 5.5. LAWS OF EXPONENTS 343 5.5 Laws of Exponents In Chapter 1, section 1, we first introduced the definition of an exponent. For convenience, we repeat that definition. In the exponential expression a™, the number a is called the base, while the number n is called the exponent. Exponents. Let a be any real number and let n be any whole number. If ra^O, then: a n = a ■ a ■ a a That is, to calculate a", write a as a factor n times. In the case where a/0, but n = 0, then we define: o° = l For example, raising a number to the fifth power requires that we repeat the number as a factor five times (see Figure 5.35). (_2) 5 = (-2)(-2)(-2)(-2)(-2) = -32 Raising a number to the fourth power requires that we repeat that number as a factor four times (see Figure 5.35). 1 H H H i 16 As a final example, note that 10° = 1, but 0° is undefined (see Figure 5.36). <-l/-2) A 4 RnsN-Frac -32 .0625 1/16 Figure 5.35: Evaluating (—2) and (-1/2) 4 . Recall: Use the MATH key to locate the ►Frac command. 10^0 0^0 1 Error Figure 5.36: Evaluating 10° and 0° on the graphing calculator. 344 CHAPTER 5. POLYNOMIAL FUNCTIONS For those who may be wondering why a = 1, provided a ^ 0, here is a nice argument. First, note that a-a° = a 1 -a° On the right, repeat the base and add the exponents. a ■ a = a Or equivalently: Multiplying With Like Bases In the expression a n , the number a is called the base and the number n is called the exponent. Frequently, we'll be required to multiply two exponential expressions with like bases, such as x 3 • x 4 . Recall that the exponent tells us how many times to write each base as a factor, so we can write: x 3 ■ x = (x ■ x ■ x) ■ (x ■ x ■ x ■ x) tJU *Xj tJU Ju Ju t*> iA; = x 7 Note that we are simply counting the number of times that x occurs as a factor. First, we have three it's, then four cc's, for a total of seven x J s. However a little thought tells us that it is much quicker to simply add the exponents to reveal the total number of times x occurs as a factor. a ■ a = a 3 4 X ■ X „3+4 Now, divide both sides by a, which is permissable if a 7^ 0. a ■ a a a a Simplify both sides: The preceding discussion is an example of the following general law of expo- nents. Multiplying With Like Bases. To multiply two exponential expressions with like bases , repeat the base and add the exponents. m a n m-\-n a = a You Try It! Simplify: 3 4 • 3 2 EXAMPLE 1. Simplify each of the following expressions: (a)y 4 V (b)2 3 -2 5 {c)ix + y) 2 {x + y y Solution: In each example we have like bases. Thus, the approach will be the same for each example: repeat the base and add the exponents. (a) y A ■ y s = y i+s (b) 2 3 • 2 5 = 2 3 + 5 (c) (x + yf{x + y) 7 = (x + yf +7 Answer: 3 6 y 12 {x + yf With a little practice, each of the examples can be simplified mentally. Repeat the base and add the exponents in your head: y ■ y (x + y) 2 (x + y) 7 = {x + yf. V 2 A ■ 2 b 2 8 , and □ 5.5. LAWS OF EXPONENTS 345 You Try It! EXAMPLE 2. Simplify: (a 6 & 4 )(a 3 6 2 ) Solution: We'll use the commutative and associative properties to change the order of operation, then repeat the common bases and add the exponents. Simplify: (x y )(x y ) 3(,2n (a b b 4 ){a 3 b a 6 b 4 a 3 b 2 a 6 a 3 b 4 b 2 9 7,6 a y 6 The associative property allows us to regroup in the order we prefer. The commutative property allows us to change the order of multiplication. Repeat the common bases and add the exponents. With practice, we realize that if all of the operators are multiplication, then we can multiply in the order we prefer, repeating the common bases and adding the exponents mentally: (a b )(a ,3/,2n. J 6 6 . Answer: x 6 y 9 □ You Try It! EXAMPLE 3. Simplify: x n+3 ■ x 3 ~ 2n Solution: Again, we repeat the base and add the exponents. ; (n+3)+(3-2n) R epea t the base, add the exponents. Simplify: x 5—n , 4n+2 x n+3 ■ x 3 ~ 2n „6— n Simplify. Combine like terms. Answer: x 3n+r Dividing With Like Bases Like multiplication, we will also be frequently asked to divide exponential ex- pressions with like bases, such as x 7 /x A . Again, the key is to remember that the exponent tells us how many times to write the base as a factor, so we can write: □ X t*> iXj *Xj *AJ iJU *JU X x ■ x ■ x ■ x- r p ■ ht ' It ' T* ■ If* ■ If* x'-x'-x'-x'- Note how we cancel four x's in the numerator for four x's in the denominator. However, in a sense we are "subtracting four x's" from the numerator, so a 346 CHAPTER 5. POLYNOMIAL FUNCTIONS Here is another nice argument why a° = 1, provided a/0. Start with: faster way to proceed is to repeat the base and subtract the exponents, as follows: „7-4 1 Repeat the base and subtract the exponents. a 1 - 1 = 1 Simplify. The preceding discussion is an example of the second general law of expo- nents. Dividing With Like Bases. To divide two exponential expressions with like bases, repeat the base and subtract the exponents. Given a/0, You Try It! 4 5 Simplify: — Note that the subtraction of the exponents follows the rule "top minus bottom." EXAMPLE 4. Simplify each of the following expressions: ™12 X 6 (b)lf (c) (2x+l) 8 (2a; + 1) 3 Solution: In each example we have like bases. Thus, the approach will be the same for each example: repeat the base and subtract the exponents. ( a )^T „12-3 5 7 - 7 5° 1 (c) (2x + 1) 8 (2x + 1) 3 (2s + 1) (2s + 1) 8-3 Answer: 4 2 With a little practice, each of the examples can be simplified mentally. Repeat the base and subtract the exponents in your head: x 12 /x 3 = x 9 , 5 7 /5 4 = 5 3 , and (2x + l) 8 /(2ir + l) 3 = (2x + l) 5 . D You Try It! Simplify: 15a 6 6 9 3a6 5 5, ,7 EXAMPLE 5. Simplify: t X „\ Solution: We first express the fraction as a product of three fractions, the latter two with a common base. In the first line of the following solution, 5.5. LAWS OF EXPONENTS 347 note that if you multiply numerators and denominators of the three separate fractions, the product equals the original fraction on the left. 5„,7 12 x 5 y 7 j.3 y2 I2x*y 4 X 3y2 4 = 3x 5 - V- 2 = 3xV Break into a product of three fractions. Simplify: 12/4=3. Then repeat the common bases and subtract the exponents. Simplify. Answer: 5a 5 6 4 □ You Try It! „5n— 4 EXAMPLE 6. Simplify: »3— 2n Solution: Again, we repeat the base and subtract the exponents. „5n— 4 Simplify: ~3n— 6 „n+2 ;)'■ 3-2n „(5n-4)-(3-2n) „5n-4-3+2ra „7n-7 Repeat the base, subtract exponents. Distribute the minus sign. Simplify. Combine like terms. Answer: n 2n-8 Raising a Power to a Power Suppose we have an exponential expression raised to a second power, such as (a; 2 ) 3 . The second exponent tells us to write x 2 as a factor three times: l 2\3 2 2 (X ) = X ■ X ■ X „6 Write x as a factor three times. Repeat the base, add the exponents. Note how we added 2 + 2 + 2 to get 6. However, a much faster way to add "three twos" is to multiply: 3-2 = 6. Thus, when raising a "power to a second power," repeat the base and multiply the exponents, as follows: (x 2 ) 3 = x 2 - 3 ™6 □ The preceding discussion gives rise to the following third law of exponents. 348 CHAPTER 5. POLYNOMIAL FUNCTIONS Answer: a 6— 3n Raising a Power to a Power. When raising a power to a power, repeat the base and multiply the exponents . In symbols: (a m ) n = a mn Note that juxtaposing two variables, as in inn, means 'm times n.' 5 You Try It! Simplify: (2 3\4 EXAMPLE 7. Simplify each of the following expressions: (a) (z 3 ) 5 (b) (7 3 ) (c) [(a; - y) 3 ] 6 Solution: In each example we are raising a power to a power. Hence, in each case, we repeat the base and multiply the exponents. 3\5 _ „3-5 (a) (zy = z (b) (7 3\0 _ 7 3-0 (c) [(x - y) 3 ifi v lo (x-yf 6 (x-y) 18 Answer: 2 12 With a little practice, each of the examples can be simplified mentally. Repeat the base and multiply the exponents in your head: (z and [(x — y) 3 ] 6 = (x — y) ls - 3\5 _ ,15 (^4 (7 3 7 12 □ You Try It! Simplify: (a 2-n\3 „2n-3\4 EXAMPLE 8. Simplify: Solution: Again, we repeat the base and multiply the exponents. (,: 2n-3\4 _ 4(2n-3) Repeat the base, multiply exponents. Distribute the 4. □ 5.5. LAWS OF EXPONENTS 349 Raising a Product to a Power We frequently have need to raise a product to a power, such as {xy) 3 . Again, remember the exponent is telling us to write xy as a factor three times, so: to/) 3 {xy){xy){xy) xyxyxy xxxyyy x 3 y 3 Write xy as a factor three times. The associative property allows us to group as we please. The commutative property allows us to change the order as we please. Invoke the exponent definition: xxx = x and yyy = y . However, it is much simpler to note that when you raise a product to a power, you raise each factor to that power. In symbols: {xy? x 3 y 3 The preceding discussion leads us to a fourth law of exponents. Raising a P roduct to a Power. To raise a product to a power, raise each factor to that power. In symbols: (ab) n = --a n b n EXAMPLE 9. Simplify each of the following expressions: (a) {yzf (b) {-2xf (c) (-3y) 2 Solution: In each example we are raising a product to a power. Hence, in each case, we raise each factor to that power. You Try It! Simplify: {-2b) 4 5,5 (a) {yzf = y b z (b) (-2a;) 3 = (-2) 3 a; 3 = -8x 3 (c) {-3yf = (-3)V = 9y 2 With a little practice, each of the examples can be simplified mentally. Raise each factor to the indicated power in your head: {yzf = y 5 z 5 , (—2a;) 3 = —8a; 3 , and (-3y) 2 = 9y 2 Answer: 166 When raising a product of three factors to a power, it is easy to show that we should raise each factor to the indicated power. For example, (abcf 3 6 3 c In general, this is true regardless of the number of factors. When raising a product to a power, raise each of the factors to the indicated power. □ 350 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! Simplify: (—3xy 4\5 EXAMPLE 10. Simplify: (-2a 3 6 2 ) 3 Solution: Raise each factor to the third power, then simplify. (-2a 3 6 2 ) 3 = (-2) 3 (a 3 ) 3 (6 2 ) 3 Raise each factor to the third power. = — 8a 9 & 6 Simplify: (— 2) 3 = 8. In the remaining factors, raising a power to a power requires that we multiply the exponents. Answer: — 243xy 5, ,20 □ You Try It! Simplify: (—a 3/,2\3/ -2a 2 b 2ui\2 Answer: — 4o b 131,14 EXAMPLE 11. Simplify: (-2x 2 y) 2 {-3x 3 y) Solution: In the first grouped product, raise each factor to the second power. (-2x 2 y) 2 (-3x 3 y) = ((-2) 2 (x 2 ) 2 y 2 )(-3x 3 y) (Ax A y 2 )(-3x 3 y) Raise each factor in the first grouped product to the second power. Simplify: (x 2 ) 2 =x 4 and The associative and commutative property allows us to multiply all six factors in the order that we please. Hence, we'll multiply 4 and —3, then x 4 and x 3 , and y 2 and y, in that order. In this case, we repeat the base and add the exponents. -12xV Simplify: (4)(-3) = x 7 and y 2 y x 4 x 3 -12. Also, .,3 □ Raising a Quotient to a Power Raising a quotient to a power is similar to raising a product to a power. For example, raising (x/y) 3 requires that we write x/y as a factor three times. \ 3 x I) X X X y y y X X ■ X vvv ^■3 5.5. LAWS OF EXPONENTS 351 However, it is much simpler to realize that when you raise a quotient to a power, you raise both numerator and denominator to that power. In symbols: x\ x 3 ,yj V 3 This leads to the fifth and final law of exponents. Raising a Quotient to a Power. To raise a quotient to a power, raise both numerator and denominator to that power. Given 6^0, (a\ n a n \b ) ~ 6" EXAMPLE 12. Simplify each of the following expressions: ' : >>(^) 2 (V(~Y (c) Solution: In each example we are raising a quotient to a power. Hence, in each case, we raise both numerator and denominator to that power. You Try It! Simplify: ""!§ 22 3 2 4 9 <«(f)H X 3 ~ 27 (c) 24 16 Note that in example (c), raising a negative base to an even power produces a positive result. With a little practice, each of the examples can be simplified mentally. Raise numerator and denominator to the indicated power in your head: (2/3) 2 = 4/9, (x/3) 3 = £ 3 /27, and (-2/j/) 4 = 16/t/ 4 Answer: 125 "64" □ EXAMPLE 13. Simplify: I — 5- V V Solution: Raise both numerator and denominator to the second power, then simplify. 2 '~„5\2 2a; 5 y 3 (2x 5 f (y 3 ) 2 Raise numerator and denominator to the second power. You Try It! Simplify: ( — ^ Answer: ^ 352 CHAPTER 5. POLYNOMIAL FUNCTIONS In the numerator, we need to raise each factor of the product to the second power. Then we need to remind ourselves that when we raise a power to a power, we multiply the exponents. 2/„.5\2 2 2 (x 5 ) (y 3 ) 2 Raise each factor in the numerator to the second power l2 a l„b\2 _ 10 Simplify: 2 Z =4, (x"Y = x l V and (y 3 ) 2 = y e . D 5.5. LAWS OF EXPONENTS 353 **. **. **. Exercises -*s -« >•$ In Exercises 1-8, simplify each of the given exponential expressions. 1. (-4) 3 2. (-9) 2 o 3. 4. 5 o 5. B) 2 6. (-!)" 7. (-19)0 8. (-17)0 In Exercises 9-18, simplify each of the given exponential expressions. 9. (7v - 6w) 18 ■ (7v - 6w) 17 14. 4 6m + 5 • 4 m ~ 5 10. (8a + 7c) 3 - (8a + 7c) 19 15. a; 8 ■ x 3 11. 3 4 - 3° 16. a 9 -a 15 12. 5 7 - 5° 17. 2 5 -2 3 13. 4" • 4 8 "+ 3 18. 2 10 ■ 2 3 In Exercises 19-28, simplify each of the given exponential expressions. 4 16 nA (46 + 7c) 15 19 416 3 12 20. — 3 12 21.— to' c io 22. — r (9a -8c) 15 ' (9a-8c)8 (46 + 7c) 5 29n+5 25. 23n-4 24fc-9 26. 23fe-8 27. 4 17 49 28. 2 17 2 6 29. (4 8m ~ 6 )' 30. (2 2m " 9\ 3 31. [(to + 52/) 3 ] 7 32. [(4u- -) 8 ] 9 33. (43) 2 354 CHAPTER 5. POLYNOMIAL FUNCTIONS In Exercises 29-38, simplify each of the given exponential expressions. 34. (3 4 ) 2 35. ( c y 36. (w 9 ) 5 37. (6 2 )° 38. (8 9 )° In Exercises 39-48, simplify each of the given exponential expressions. 39. (uw) 5 44. (-3w 9 ) 4 40. {ac) 4 45. (-3x s y 2 ) 4 41. (-2y) 3 46. (2x 8 z 6 ) 4 42. {-2b) 3 47. (7s 6 ™) 3 43. (3w 9 ) 4 48. (96 6 ") 3 In Exercises 49-56, simplify each of the given exponential expressions. 49. (- V2 . 2^ 2 51. u / q \ 3 52. I-- 53. (4) 4 54. (-£)" 55. (I)* 56. \u 12 5.5. LAWS OF EXPONENTS 355 57. Complete each of the laws of exponents presented in the following box, then use the results to simplify the expressions to the right of the box. a m a n ? a m ? a n (a m ) n = ? (ab) m = ? (!)" = 7 a 3 a 5 ,5\7 _ = ? 2 (« b ) 7 = ? (a&) 9 = ? «*» ? m- j*- j*- Answers •*$ •** •** 1. -64 3. 1 9 7. 1 9. (7w - 6w) 35 11. 3 4 13. 4 9n+3 15. a; 11 17. 2 8 19. 1 21. w 4 23. (9a -8c) 7 25. 2 6n+9 27. 4 8 OQ /|56m— 42 31. (9s + 5y) 21 33. 4 6 35. c 28 37. 1 39. u 5 w 5 41. -8y 3 43. 81w 36 45. 81x 32 y s 47. 343s 18 " v 49. — 51. 53. 55. 57. 625 625 c 36 25 » 21 356 CHAPTER 5. POLYNOMIAL FUNCTIONS Answer: -18s 4 i 6 5.6 Multiplying Polynomials In this section we will find the products of polynomial expressions and func- tions. We start with the product of two monomials, then graduate to the product of a monomial and polynomial, and complete the study by finding the product of any two polynomials. The Product of Monomials As long as all of the operations are multiplication, we can use the commutative and associative properties to change the order and regroup. You Try It! Multiply: 3 (Ax) EXAMPLE 1. Multiply: -5(7j/). Simplify: -5(7y). Solution: Use the commutative and associative properties to change the order and regroup. -5(7y) [(-5)(7)]y -35y Reorder. Regroup. Multiply: (-5) (7) : -35. Answer: 12a; □ You Try It! Multiply: (7y 5 )(-2y 2 EXAMPLE 2. Multiply: (-3a; 2 ) (4a; 3 ). Simplify: (-3a; 2 ) (4a; 3 ). Solution: Use the commutative and associative properties to change the order and regroup. (-3a; 2 )(4a; 3 ) = [(-3)(4)](xV) Reorder. Regroup. = -12a; 5 Multiply: (-3) (4) 12, a; 2 x 3 =x 5 . Answer: -Uy 7 □ You Try It! Multiply: (-6st 2 )(3s 3 t 3+4\ EXAMPLE 3. Multiply: (-2a 2 6 3 )(-5a 3 6). Simplify: (-2a 2 6 3 )(-5a 3 6). Solution: Use the commutative and associative properties to change the order and regroup. (-2a 2 6 3 )(-5a 3 6) = [(-2)(-5)](aV)(6 3 6) Reorder. Regroup 10a 5 6 4 Multiply: (-2)(-5) = 10, a 2 a 3 a b , and b A b = 6 4 □ 5.6. MULTIPLYING POLYNOMIALS 357 When multiplying monomials, it is much more efficient to make the required calculations mentally. In the case of Example 1, multiply —5 and 7 mentally to get -5(7y) = -35y. In the case of Example 2, multiply —3 and 4 to get —12, then repeat the base x and add exponents to get (-3a; 2 )(4a; 3 -12a; a . Finally, in the case of Example 3, multiply —2 and —5 to get 10, then repeat the bases and add their exponents. ■i;,4 {-2a% 6 ){-ha 6 b) = 10a b 6 Multiplying a Monomial and a Polynomial Now we turn our attention to the product of a monomial and a polynomial. EXAMPLE 4. Multiply: 5(3a: 2 - Ax - 8) Solution: We need to first distribute the 5 times each term of the polynomial. Then we multiply the resulting monomials mentally. 5(3x 2 - 4a; - 8) = 5(3a; 2 ) - 5(4ar) - 5(8) = 15a; 2 - 20a; - 40 You Try It! Multiply: A(-2y 2 + 3y + 5) Answer: -8y 2 + 12y + 20 □ EXAMPLE 5. Multiply: 2j/(-3y + 5) Solution: We need to first distribute the 2y times each term of the polynomial. Then we multiply the resulting monomials mentally. 2y(-3y + 5) = 2y(-3y) + 2y(5) = -6y 2 + lOy You Try It! Multiply: -3a: (4 - 2x) Answer: — 12a; + 6a; 2 □ 358 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! Multiply: 2xy(x 2 - 4xy 2 + 7x) EXAMPLE 6. Multiply: -3a6(a 2 + 2ab - b 2 ) Solution: We need to first distribute the — 3ab times each term of the poly- nomial. Then we multiply the resulting monomials mentally. -3ab(a 2 + 2ab - b 2 ) -3ab(a 2 ) + (-3ab)(2ab) - (-3ab){b 2 ) -3a 3 b + (-6a 2 b 2 ) - (Sab 3 ) -3a 3 b - 6a 2 b 2 + 3ab 3 Answer: 2x 3 y - 8x 2 y 3 + \Ax 2 y Alternate solution: It is far more efficient to perform most of the steps of the product — 3a6(a 2 + 2ab — b 2 ) mentally. We know we must multiply — 3ab times each of the terms of the polynomial a 2 + 2ab — b 2 . Here are our mental calculations: i) — 3ab times a 2 equals — 3a 3 b. ii) — 3ab times 2ab equals — 6a 2 b 2 . iii) — 3ab times — b 2 equals 3ab 3 . Thinking in this manner allows us to write down the answer without any of the steps shown in our first solution. That is, we immediately write: -3ab(a 2 + 2ab- b 2 -3a 3 b-6a 2 b 2 + 3ab 3 D You Try It! Multiply: -5y 3 {y 2 - 2y + 1 EXAMPLE 7. Multiply: -2z 2 {z 3 + Az 2 - 11) Solution: We need to first distribute the — 2z 2 times each term of the poly- nomial. Then we multiply the resulting monomials mentally. -2z 2 {z 3 + 4z 2 11) -2z 2 (z 3 ) + (-2z 2 ){Az 2 ) - (-2z 2 )(ll) -2z 5 + (-8z 4 )-(-22z 2 ) -2z b iz A + 22z 2 Alternate solution: It is far more efficient to perform most of the steps of the product — 2z 2 (z 3 + 4z 2 — 11) mentally. We know we must multiply — 2z 2 times each of the terms of the polynomial z 3 + 4z 2 — 11. Here are our mental calculations: 5.6. MULTIPLYING POLYNOMIALS 359 i) — 2z 2 times z 3 equals — 2z 5 . ii) — 2z 2 times 4z 2 equals — 8z 4 . iii) — 2z 2 times —11 equals 22z 2 . Thinking in this manner allows us to write down the answer without any of the steps shown in our first solution. That is, we immediately write: -2z 2 (z 3 + 4z 2 - 11) = -2z 5 - 8z 4 + 22z 2 Answer: — by 5 + 10y 4 — 5y 3 □ Multiplying Polynomials Now that we've learned how to take the product of two monomials or the product of a monomial and a polynomial, we can apply what we've learned to multiply two arbitrary polynomials. Before we begin with the examples, let's revisit the distributive property. We know that 2- (3 + 4) = 2-3 + 2-4. Both sides are equal to 14. We're used to having the monomial factor on the left, but it can also appear on the right. (3 + 4) -2 = 3-2 + 4-2 Again, both sides equal 14. So, whether the monomial appears on the left or right makes no difference; that is, a(b + c) = ab + ac and (b + c)a = ba + ca give the same result. In each case, you multiply a times both terms of b + c. EXAMPLE 8. Multiply: (2x + 5) (3a; + 2) Solution: Note that (2x + 5) (3a; + 2) has the form (b + c)a, where a is the binomial 3a; + 2. Because (b + c)a = ba + ca, we will multiply 3a; + 2 times both terms of 2a; + 5 in the following manner. (2a; + 5)(3a; + 2) = 2a;(3a; + 2) + 5(3a; + 2) Now we distribute monomials times polynomials, then combine like terms. You Try It! Multiply: (3.x + 4) (5a 6x z ix + 15a; + 10 Thus, (2a; + 5)(3a; + 2) = 6x 2 + 19a; + 10 6a; 2 + 19a; + 10. Answer: 15a; 2 + 23a; + 4 □ 360 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! Multiply: (z + 4)(z 2 + 3z + 9) EXAMPLE 9. Multiply: (x + 5) (a; 2 + 2x + 7) Solution: Note that (a; + 5) (a; 2 + 2x + 7) has the form (b + c)a, where a is the binomial 3a; + 2. Because (6 + c)« = ba + ca, we will multiply a; 2 + 2a; + 7 times both terms of x + 5 in the following manner. (a; + 5)(.t 2 + 2x + 7) = x(x 2 + 2a; + 7) + 5(a; 2 + 2a; + 7) Now we distribute monomials times polynomials, then combine like terms. = x 3 + 2x 2 + 7x + 5x 2 + Wx + 35 Answer: z 3 + 7z 2 + 21 z + 36 = x 3 + 7x 2 + YIx + 35 Thus, (a; + 5)(.x 2 + 2a; + 7) = a; 3 + 7a; 2 + 17a; + 35. □ Speeding Things Up a Bit Let's rexamine Example 9 with the hope of unearthing a pattern that will allow multiplication of polynomials to proceed more quickly with less work. Note the first step of Example 9. (x + 5){x 2 + 2x + 7) = a;(a; 2 + 2a; + 7) + 5(a; 2 + 2a; + 7) Note that the first product on the right is the result of taking the product of the first term of x + 5 and x 2 + 2x + 7. Similarly, the second product on the right is the result of taking the product of the second term of x + 5 and a; 2 + 2a; + 7. Next, let's examine the result of the second step. (x + 5)(x 2 + 2x + 7) = x 3 + 2x 2 + 7x + 5x 2 + Wx + 35 The first three ter ms on the right are t he result of multiplying x times a; 2 +2x+7. 2x + 7) 1 {x + 5)(x' + 2x + 7) 2x A •7ar+' The second set of three terms on the right are the result of multiplying 5 times a; 2 + 2a; + 7. I J I 1 5)(a; 2 + 2a; + 7) = - (a; + 5)(a 5a;^ 10a; + 35 These notes suggest an efficient shortcut. To multiply x + 5 times a; 2 + 2a; + 7, • multiply x times each term of x 2 + 2x + 7, then • multiply 5 times each term of a; 2 + 2a; + 7. • Combine like terms. 5.6. MULTIPLYING POLYNOMIALS 361 This process would have the following appearance. (a; + 5)(a; 2 + 2x + 7)=x 3 + 2x 2 + 7x + 5a; 2 + Wx + 35 17a; + 35 x 3 + 7x 2 You Try It! EXAMPLE 10. Use the shortcut technique described above to simplify Multiply: (3a; + 2){Ax 2 - x + 10) (2y-6)(3 2 / 2 + 4 2 /+ll) Solution: Multiply 2y times each term of 3j/ 2 + Ay + 11, then multiply —6 times each term of 3j/ 2 + Ay + 11. Finally, combine like terms. (2y - 6){3y 2 + Ay + 11) = 6y 3 + 8y 2 + 22y - 18y 2 - 2Ay - 66 = 6y 3 - 10y 2 -2y-66 Answer: 12z 3 + 5a; 2 + 28a; + 20 D EXAMPLE 11. Use the shortcut technique to simplify (a + b) 2 . Solution: To simplify (a + b) 2 , we first must write a + b as a factor two times. {a + bf = (a + b)(a + b) Next, multiply a times both terms of a + b, multiply b times both terms of a + b, then combine like terms. = a 2 + ab + ba + b 2 = a 2 + 2ab + b 2 Note that ab = ba because multiplication is commutative, so ab + ba = 2ab. You Try It! Multiply: (3y - 2) Answer: 9y 2 — 12y + A □ EXAMPLE 12. Use the shortcut technique to simplify (x 2 +x + l)(x 2 - x-l). You Try It! Multiply: (a 2 -2a + 3)(a 2 + 2a-3) 362 CHAPTER 5. POLYNOMIAL FUNCTIONS Solution: This time the first factor contains three terms, x 2 , x, and 1, so we first multiply x times each term of x x 2 — x — 1, and 1 times each term of x 2 — (x 2 + x + \){x 2 - x - 1) = x 4 - x 3 - 1, then x times each term of x - 1 Then we combine like terms. x 2 + x 3 - x 2 - x + x 2 — x — 1 2x — 1 Answer: a 4 - 4a 2 + 12a - 9 a Some Applications Recall that the area of a circle of radius r is found using the formula A = irr 2 . The circumference (distance around) of a circle of radius r is found using the formula C = 2irr (see Figure 5.37). Figure 5.37: Formulae for the area and circumference of a circle of radius r. You Try It! Two concentric circles are shown below. The inner circle has radius x and the outer circle has radius x + 2. Find the area of the shaded region as a function of x. EXAMPLE 13. In Figure 5.38 are pictured two concentric circles (same center). The inner circle has radius x and the outer circle has radius x + I. Find the area of the shaded region (called an annulus) as a function of x. Figure 5.38: Find the area of the shaded region. 5.6. MULTIPLYING POLYNOMIALS 363 Solution: We can find the area of the shaded region by subtracting the area of the inner circle from the area of the outer circle. A = Area of outer circle — Area of inner circle We use the formula A = irr 2 to compute the area of each circle. Because the outer circle has radius x + 1, it has area w(x + l) 2 . Because the inner circle has radius x, it has area irx 2 . = ir(x + 1) — irx Next, we'll expand (x + l) 2 , then combine like terms. = ir(x + l)(x + 1) — irx = n(x 2 + x + x + 1) — ttx 2 = ir(x 2 + 2x + 1) - ttx 2 Finally, distribute tt times each term of x + 2x + 1, then combine like terms. = TTX + 2lTX + TV — TTX = 2wX + TT Hence, the area of the shaded region is A = 2ttx + tt. Answer: Anx + An □ You Try It! EXAMPLE 14. The demand for widgets is a function of the unit price, where the demand is the number of widgets the public will buy and the unit price is the amount charged for a single widget. Suppose that the demand is given by the function x = 270 — 0.75p, where x is the demand and p is the unit price. Note how the demand decreases as the unit price goes up (makes sense). Use the graphing calculator to determine the unit price a retailer should charge for widgets so that his revenue from sales equals $20,000. Solution: To determine the revenue (R), you multiply the number of widgets sold (x) by the unit price (p). R = xp (5.5) However, we know that the number of units sold is the demand, given by the formula x = 270-0.75p (5.6) Substitute equation 5.6 into equation 5.5 to obtain the revenue as a function of the unit price. R = (270 - 0.75p)p Suppose that the demand for gadgets is given by the function x = 320 — 1.5p, where x is the demand and p is the unit price. Use the graphing calculator to determine the unit price a retailer should charge for gadgets so that her revenue from sales equals $12,000. 364 CHAPTER 5. POLYNOMIAL FUNCTIONS Expand. In this example, the horizontal axis is actually the p-axis. So when we set Xmin and Xmax, we're actually setting bounds on the p-axis. R = 270p - 0.75p 2 (5.7) We're asked to determine the unit price that brings in a revenue of $20,000. Substitute $20,000 for R in equation 5.7. 20000 = 270p - 0.75]/ (5.8) Enter each side of equation 5.8 into the Y= menu of your calculator (see the first image in Figure 5.39). After some experimentation, we settled on the WINDOW parameters shown in the second image of Figure 5.39. After making these settings, push the GRAPH button to produce the graph shown in the third image in Figure 5.39. Plotl PlotE Plots WiB270*X-0.75*X '-2 W £020000 sVs = WINDOW Xnin=0 Xnax=400 Xscl=100 Ymin=0 Vnax=30000 Y£Cl=10000 ■i-Xres=l Figure 5.39: Solving 20000 = 270p- 0.75p 2 . Note that the third image in Figure 5.39 shows that there are two solutions, that is, two ways we can set the unit price to obtain a revenue of $20,000. To find the first solution, select 5:intersect from the CALC menu, press ENTER in response to "First curve," press ENTER in response to "Second curve," then move your cursor closer to the point of intersection on the left and press ENTER in response to "Guess." The result is shown in the first image in Figure 5.40. Repeat the process to find the second point of intersection. The result is shown in the second image in Figure 5.40. Intersection K=i04.£Bl££ ..Y=£0000. Intersection K=2££.7iB7H ..Y=£0000 . Figure 5.40: Finding the points of intersection. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with p and R, respectively (see Figure 5.41). Include the units (dollars and dollars). Place your WINDOW parameters at the end of each axis (see Figure 5.41). 5.6. MULTIPLYING POLYNOMIALS 365 30000 , _R(dollars) R = 20000 R = 270p - 0.75p 2 p(dollars) 104.28122 255.71878 400 Figure 5.41: Reporting your graphical solution on your homework. • Label each graph with its equation (see Figure 5.41). • Drop a dashed vertical line through each point of intersection. Shade and label the p- values of the points where the dashed vertical lines cross the p-axis. These are the solutions of the equation 20000 = 270p — 0.75p 2 (see Figure 5.41). Rounding to the nearest penny, setting the unit price at either $104.28 or $255.72 will bring in a revenue of $20,000. Answer: .55 or $164.79 □ 366 CHAPTER 5. POLYNOMIAL FUNCTIONS ;».;-». t± Exercises •*;■*; -*i In Exercises 1-10, simplify the given expression. 1. -3(7r) 2. 7(3a) 3. (-96 3 )(-86 6 ) 4. (8s 3 )(-7s 4 ) 5. (-7r 2 £ 4 )(7r 5 £ 2 ) 6. (-10s 2 t 8 )(-7s 4 i 3 7. (-56 2 c 9 )(-86 4 c 4 ) 8. (-9s 2 i 8 )(7s 5 t 4 ) 9. {-8v 3 )(4v 4 ) 0. (-9y 3 )(3y 5 ) In Exercises 11-22, use the distributive property to expand the given expression. 11. 9(-26 2 + 26 + 9) 12. 9(-46 2 + 76-8) 13. -4(10i 2 -7i-6) 14. -5(-7w 2 -7u + 2) 15. -8u 2 (-7u 3 - 8m 2 -2u + 10) 16. -3s 2 (-7s 3 - 9s 2 + 6s + 3) 17. 10s 2 (-10s 3 + 2s 2 + 2s + 8 18. 8w 2 (9u 3 -5u 2 -2u + 5) 19. 2st(-4s 2 + 8st - lQt 2 ) 20. 7uv(-9u 2 - 3uv + 4v 2 ) 21. -2uu;(10w 2 - 7uw - 2w 2 ) 22. -6vw(-5v 2 + 9vw + 5w 2 ) In Exercises 23-30, use the technique demonstrated in Examples 8 and 9 to expand each of the following expressions using the distributive property. 23. (-9x-4)(-3x + 2) 24. (4x- 10)(-2a:-6) 25. (3a; + 8)(3a;-2) 26. (-6x+8)(-x+l) 27. (2x- l)(-6x 2 +4x + 5) 28. (Ax - 6)(-7ir 2 - lOcc + 10) 29. (x-6)(-2x 2 -Ax -4) 30. (5x- 10)(-3a; 2 + 7a;-8) In Exercises 31-50, use the shortcut technique demonstrated in Examples 10, 11, and 12 to expand each of the following expressions using the distributive property. 31. (8u - 9w)(8u - 9w) 32. (36 + 4c)(-86+10c) 33. (9r- 7t)(3r-9t) 34. (-6x-3y)(-6x + 9y) 5.6. MULTIPLYING POLYNOMIALS 367 35. {At - 10s)(-10r 2 + lOrs - 7s 2 ) 36. (5s - 9f)(-3s 2 + Ast - 9t 2 ) 37. (9a; - 2z)(4x 2 - Axz - Wz 2 ) 38. (r - 4i)(7r 2 + Art - 2t 2 ) 39. (9r + 3i) 2 40. (4a; + 8z) 2 41. (Ay + 5z)(Ay - 5z) 42. (7w + 2w)(7w-2w) 43. (7u + 8v)(7u - 8v) 44. (66 + 8c)(66-8c) 45. (76 + 8c) 2 46. (26 + 9c) 2 47. (2i 2 + 9t + A)(2t 2 + 9t + A) 48. (3a 2 - 9a + 4)(3a 2 - 9a + 2) 49. (Aw 2 + 3w + 5)(3w 2 - 6w + 8) 50. (4s 2 + 3s + 8)(2s 2 + 4s - 9) 51. The demand for widgets is given by the function x = 320 — 0.95p, where x is the demand and p is the unit price. What unit price should a retailer charge for widgets in order that his revenue from sales equals $7,804 ? Round your answers to the near- est cent. 52. The demand for widgets is given by the function x = 289 — 0.91p, where x is the demand and p is the unit price. What unit price should a retailer charge for widgets in order that his revenue from sales equals $7,257 ? Round your answers to the near- est cent. 53. In the image that follows, the edge of the outer square is 6 inches longer than 3 times the edge of the inner square. 54. In the image that follows, the edge of the outer square is 3 inches longer than 2 times the edge of the inner square. □ Express the area of the shaded region as a polynomial in terms of x, the edge of the inner square. Your final answer must be presented as a second degree polynomial in the form A(x) bx - b) Given that the edge of the inner square is 5 inches, use the polynomial in part (a) to determine the area of the shaded region. Express the area of the shaded region as a polynomial in terms of x, the edge of the inner square. Your final answer must be presented as a second degree polynomial in the form A(x) bx + c. b) Given that the edge of the inner square is 4 inches, use the polynomial in part (a) to determine the area of the shaded region. 368 CHAPTER 5. POLYNOMIAL FUNCTIONS 55. A rectangular garden is surrounded by a uniform border of lawn measuring x units wide. The entire rectangular plot mea- sures 31 by 29 feet. 29 X 31 Find the area of the interior rectangu- lar garden as a polynomial in terms of x. Your final answer must be presented as a second degree polynomial in the form A(x) bx + c. b) Given that the width of the border is 9.3 feet, use the polynomial in part (a) to de- termine the area of the interior rectangular garden. 56. A rectangular garden is surrounded by a uniform border of lawn measuring x units wide. The entire rectangular plot mea- sures 35 by 24 feet. 24 X 35 a) Find the area of the interior rectangu- lar garden as a polynomial in terms of x. Your final answer must be presented as a second degree polynomial in the form A(x) = ax 2 + bx + c. b) Given that the width of the border is 1.5 feet, use the polynomial in part (a) to de- termine the area of the interior rectangular garden. j*- j*- j*- Answers •*$ **s ■** 1. -21r 3. 72b 9 5. -49r 7 * 6 7. 406 6 c 13 9. -32m 7 11. -186 2 + 186 + 81 13. -40* 2 + 28* + 24 15. 56m 5 + 64m 4 + 16m 3 - 80m 2 17. -100s 5 + 20s 4 + 20s 3 + 80s 2 19. -8s 3 * + 16s 2 * 2 - 20s* 3 21. -20u 3 w + 14u 2 w 2 + 4mm; 3 23. 27x 2 - 6a; - 8 25. 9x 2 + 18a;- 16 27. -12a; 3 + 14x 2 + 6a; - 5 29. -2a; 3 + 8a; 2 + 20a; + 24 31. 64m 2 - IUuw + 81w 2 33. 27r 2 - 102r* + 63t 2 35. -40r 3 + 140r 2 s - 128rs 2 + 70s 3 37. 36a; 3 - 44a; 2 z - 82a;z 2 + 20z 3 39. 81r 2 + 54r* + 9i 2 5.6. MULTIPLYING POLYNOMIALS 369 41. 16y 2 - 25z 2 43. 49w 2 - 64v 2 45. 496 2 + 1126c + 64c 2 47. 4f* + 36i 3 + 97£ 2 + 72t + 16 49. I2w 4 - 15w 3 + 29w 2 - 6w + 40 51. $26.47, $310.37 53. AO) = 8a; 2 + 36x + 36, A(5) = 416 square inches 55. 899 - 120a; + 4a; 2 , 128.96 square feet 370 CHAPTER 5. POLYNOMIAL FUNCTIONS 5.7 Special Products This section is dedicated to explaining a number of important shortcuts for mul- tiplying binomials. These are extremely important patterns that will produce the same products computed in previous sections. It is essential that readers practice until they become proficient using each of the patterns presented in this section. The FOIL Method Consider the product of two binomials (a; + 3)(a; + 6). We already know how to find the product of these two binomials; we multiply x times both terms of x + 6, then we multiply 3 times both terms of x + 6. (x + 3)(x + 6) = x 2 + 6x + 3a; + 18 Normally we combine like terms, but we halt the process at this point so as to introduce the pattern called the FOIL method. The letters in the word FOIL stand for "First," "Outer," "Inner," and "Last." Let's see how we can connect these terms to the product (a; + 3) (a; + 6). • The arrows indicate the terms in the "First" positions in each binomial. If you multiply the terms in the "First" position, you get x 2 . ( |+3)(| + 6) F • The arrows indicate the terms in the "Outer" positions in each binomial. If you multiply the terms in the "Outer" positions, you get 6x. ( f + 3)(s + f ) O • The arrows indicate the terms in the "Inner" positions in each binomial. If you multiply the terms in the "Inner" positions, you get 3x. (i + JKj + 6) I • The arrows indicate the terms in the "Last" positions in each binomial. If you multiply the terms in the "Last" positions, you get 18. (E + ^QE + g ) L 5.7. SPECIAL PRODUCTS 371 The following diagram shows the connection between "First," "Outer," "In- ner," "Last," and the answer. FOIL (x + 3){x + 6) = x 2 + 6x + 3a; + 18 EXAMPLE 1. Use the FOIL method to simplify: (x + 5) (a; + 7) Solution: Multiply the "First" positions: x 2 . Multiply the "Outer" positions: 7x. Multiply the "Inner" positions: 5a;. Multiply the "Last" positions: 35. F „2 o 7x (x + 5)(x + 7) -- Combining like terms, {x + 5) (a; + 7) = x 2 + 12a; + 35 I 5;); L 35 You Try It! Simplify: (x + 2)(x + 11) Answer: 13a: + 22 D EXAMPLE 2. Use the FOIL method to simplify: (2x - 7) (x - 4) Solution: Multiply the "First" positions: 2a; 2 . Multiply the "Outer" posi- tions: —8a;. Multiply the "Inner" positions: —7a;. Multiply the "Last" posi- tions: 28. You Try It! Simplify: (a;- l)(4a; + 5) (2a; -7)0 -4) Combining like terms, (2a; — 7) (a; — 4) = 2a; /•' O I L 2x 2 - 8a; - 7a; + 28 -4) = 2a; 2 - 15a; + 28. Answer: 4a; 2 + x — 5 □ At first glance, the FOIL method doesn't look like much of a shortcut. After all, if we simply use the distributive property on the product of Example 2, we get the same quick result. (2a; - 7)(x - 4) = 2a;(a; - 4) - 7{x - 4) = 2a; 2 - 8a; - 7a; + 28 = 2a; 2 - 15a; + 28 The FOIL method becomes a true shortcut when we add the "Outer" and "Inner" results in our head. 372 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! FOIL Shortcut. To multiply two binomials, follow these steps: 1. Multiply the terms in the "First" positions. 2. Multiply the terms in the "Outer" and "Inner" positions and combine the results mentally (if they are like terms) . 3. Multiply the terms in the "Last" positions. Simplify: (2z - 3)(5z - 1) EXAMPLE 3. Use the FOIL shortcut to simplify: (3x + 8) (2a - 1) Solution: Each of the following steps is performed mentally. 1. Multiply the terms in the "First" positions: 6x 2 2. Multiply the terms in the "Outer" and "Inner" positions and add the results mentally: — 3.T + 16a; = 13a; 3. Multiply the terms in the "Last" positions: —8 Write the answer with no intermediate steps: (3a; + 8) (2a; — 1) = 6a; 2 + 13a; — 8. Answer: 10z 2 - 17z + 3 □ You Try It! Simplify: (7a; + 2) (2a; - 3) Answer: 14a; 2 — 17a; — 6 EXAMPLE 4. Use the FOIL shortcut to simplify: (Ay - 3)(5y + 2) Solution: Each of the following steps is performed mentally. 1. Multiply the terms in the "First" positions: 20y 2 2. Multiply the terms in the "Outer" and "Inner" positions and add the results mentally: 8y — 15y = —7y 3. Multiply the terms in the "Last" positions: —6 Write the answer with no intermediate steps: (Ay — 3)(5y + 2) = 20j/ 2 — 7y — 6. □ 5.7. SPECIAL PRODUCTS 373 The Difference of Squares We can use the FOIL shortcut to multiply (a + 6) (a — b). 1. Multiply the terms in the "First" positions: a 2 2. Multiply the terms in the "Outer" and "Inner" positions and add the results mentally: ab — ab = 3. Multiply the terms in the "Last" positions: — b 2 Thus, (a + b)(a — b) = a 2 — b 2 . Note how the right-hand side a 2 — b 2 is the difference of two squares. This leads to the following shortcut. The difference of squares. If you have identical terms in the 'First" po- sitions and identical terms in the "Last" positions, but one set is separated with a plus sign l while the other is separated by a minus sign, then proceed as follows: 1. Square the "First" term 2. Square the "Last" term. 3. Place a minus sign between the results That is, (a- f b)(a-b) = a 2 -b 2 Note: If you don't have identical terms in the "First" and "Last" positions, with one set separated with a plus sign and the other with a minus sign, then you do not have the difference of squares pattern and you must find some other way to multiply. For example, (x + 3)(x — 3) is an example of the difference of squares pattern, but (2y + 3)(2y — 5) is not. You Try It! EXAMPLE 5. Use the difference of squares shortcut to simplify: (x + 3)(x — Simplify: (x + 5)(x — 5) 3) Solution: Note how the terms in the "First" position are identical, as are the terms in the "Last" position, with one set separated by a plus sign and the other with a minus sign. Hence, this is the difference of squares pattern and we proceed as follows: 1. Square the term in the "First" position: x 2 2. Square the term in the "Last" position: (— 3) 2 = 9 3. Separate the squares with a minus sign. 374 CHAPTER 5. POLYNOMIAL FUNCTIONS Answer: x 2 — 25 That is: (a; + 3)(x-3) = x 2 - (3) 2 9 Note: You should practice this pattern until you can go straight from the problem statement to the answer without writing down any intermediate work, as in (x + 3)(x — 3) 9. □ You Try It! Simplify: (3a - 66) (3a + 66) Answer: 9a 2 — 366 2 EXAMPLE 6. Use the difference of squares shortcut to simplify: (8y + 7z)(8y-7z) Solution: Note how the terms in the "First" position are identical, as are the terms in the "Last" position, with one set separated by a plus sign and the other with a minus sign. Hence, this is the difference of squares pattern and we proceed as follows: 1. Square the term in the "First" position: (8y) 2 = 64y 2 2. Square the term in the "Last" position: (7z) 2 = 49z 2 3. Separate the squares with a minus sign. That is: (8y + 7z)(8y-7z) = (8y) 2 -(7z) 2 = 64y 2 - 49z 2 Note: You should practice this pattern until you can go straight from the problem statement to the answer without writing down any intermediate work, as in (8y + 7z)(8y - 7z) = 64y 2 - AQz 2 . □ You Try It! Simplify: (2y A + z 3 )(2y A - z 3 ) EXAMPLE 7. Use the difference of squares shortcut to simplify: (x 3 - 5y 2 )(x 3 + by 2 ) Solution: Note how the terms in the "First" position are identical, as are the terms in the "Last" position, with one set separated by a plus sign and the other with a minus sign. Hence, this is the difference of squares pattern and we proceed as follows: 5.7. SPECIAL PRODUCTS 375 1. Square the term in the "First" position: (a; 3 ) 2 = x 6 2. Square the term in the "Last" position: (5y 2 ) 2 = 25y 4 3. Separate the squares with a minus sign. That is: (a; 3_ 5y 2 )(a; 3 +52/ 2 ) = (a; 3 ) 2_ (% 2 ) 2 = x 6 - 25y 4 Note: You should practice this pattern until you can go straight from the problem statement to the answer without writing down any intermediate work, as in (x 3 - 5y 2 )(x 3 + by 2 ) = x 6 - 25y 4 . Answer: 4y 8 - z 6 Squaring a Binomial Before demonstrating the correct procedure for squaring a binomial, we will first share one of the most common mistakes made in algebra. □ Warning! This is incorrect! One of the most common mistakes made in algebra is the assumption that: (a + bf = a 2 + b 2 The fact that this is incorrect is easily checked. Substitute 3 for a and 4 for b. (3 + 4) 2 ? 2 = 3 2 + 4 2 = 3 2 + 4 2 49 = 9 + 16 Clearly this is incorrect! So what is the correct answer? First, (a + b) 2 = (a + b)(a + b). We can now use the FOIL shortcut. 1. Multiply the terms in the "First" positions: a 2 2. Multiply the terms in the "Outer" and "Inner" positions and add the results mentally: ab + ab = 2ab 3. Multiply the terms in the "Last" positions: b 2 Hence, the correct answer is (a + b) 2 = a 2 + 2ab + b 2 . This leads us to the following shortcut for squaring a binomial. 376 CHAPTER 5. POLYNOMIAL FUNCTIONS Squaring a binomial. following steps: To square a binomial, such as (a + b) 2 , perform the 1. Square the "First" term: a 2 2. Multiply the "First" and "Last' terms and double the result: 2ab 3. Square the "Last" term: b 2 That is: (o4 bf = a 2 + 2ab + b 2 You Try It! Simplify: (a; + 3) Answer: x 2 + 6x + 9 EXAMPLE 8. Use the squaring a binomial shortcut to expand: (x + 5) 2 Solution: Follow these steps: 1. Square the first term: x 2 2. Multiply the "First" and "Last" terms and double the result: 2(x)(5) = 10a; 3. Square the "Last" term: 5 2 = 25 Thus: (a; + 5) 2 =x: 2 + 2(x:)(5) + (5) 2 = x 2 + Wx + 25 Note: You should practice this pattern until you can go straight from the problem statement to the answer without writing down any intermediate work, as in (x + 5) 2 = x 2 + Wx + 25. □ Comment. Students often refuse to learn the "squaring a binomial" shortcut, noting that they can just as easily use the FOIL technique or a simple appli- cation of the distributive property to arrive at the same result. Unfortunately, failure to learn the "squaring a binomial" shortcut will severely handicap stu- dents, as this pattern is an important component of many procedures in future mathematics courses. You Try It! Simplify: (2y + 3z) 2 EXAMPLE 9. Use the squaring a binomial shortcut to expand: (3x -I- 7y) 2 Solution: Follow these steps: 5.7. SPECIAL PRODUCTS 377 1. Square the first term: (3a:) 2 = 9x 2 2. Multiply the "First" and "Last" terms and double the result: 2(3x)(7y) = A2xy 3. Square the "Last" term: (7y) 2 = 49j/ 2 Thus: (3i + 7y) 2 = (3a;) 2 + 2(3a:)(7y) + (7y) 2 = 9a; 2 + 42a;?/ + 49j/ 2 Note: You should practice this pattern until you can go straight from the problem statement to the answer without writing down any intermediate work, as in (3a; + 7y) 2 = 9.x 2 + A2xy + 49y 2 . Answer: Ay 2 + 12yz + 9z 2 In the next example, when squaring a binomial with a minus sign, we take care of the minus sign by "adding the opposite." You Try It! EXAMPLE 10. Use the squaring a binomial shortcut to expand: (4a 2 — 5o 3 ) 2 Simplify: (3a; 4 — 5z 2 ) 2 Solution: Add the opposite: (4a 2 -56 3 ) 2 = (4a 2 + (-56 3 )) 2 . Now follow these steps: 1. Square the first term: (4a 2 ) 2 = 16a 4 2. Multiply the "First" and "Last" terms and double the result: 2(4a 2 )(-5o 3 ) = -40a 2 6 3 3. Square the "Last" term: (-56 3 ) 2 = 25b 6 Thus: (4a 2 - 56 3 ) 2 = (4a 2 + (-56 3 )) 2 = (4a 2 ) 2 + 2(4a 2 )(-56 3 ) + (-56 3 ) 2 = 16a 4 - 40a 2 6 3 + 25o 6 Note: You should practice this pattern until you can go straight from the problem statement to the answer without writing down any intermediate work, as in (4a 2 - 56 3 ) 2 = 16a 4 - 40a 2 6 3 + 25b 6 . Answer: 9a; 8 - 30a; 4 ;z 2 + 25z 4 Example 10 shows us that if we are squaring a difference, the middle term will be minus. That is, the only difference between (a + o) 2 and (a — 6) 2 is the sign of the middle term. a a 378 CHAPTER 5. POLYNOMIAL FUNCTIONS You Try It! Simplify: (a 2 - 36 5 ) 5 Answer: a 4 - 6a 2 6 5 + 96 10 Squaring a binomial. The shortcuts for squaring a binomial are: {a + bf = a 2 + 2ab + b 2 {a-bf = a 2 - 2ab + b 2 EXAMPLE 11. Use the squaring a binomial shortcut to expand: (x 3 y :?\2 Solution: Use the pattern (a — b) z 2at b 2 . Square the "First" term, multiply the "First" and "Last" terms and double the result, then square the "Last" term. Because of the minus sign, the middle term will be minus, but all other terms are plus. : '\2 {x* - V y = {x s f - 2{x i ){y i ) + (p") 2x 3 y 3 D An Application In Example 7 on page 348, we found the area of the outer square by summing the areas of its parts (see Figure 5.42). Recall that the answer was A = x 2 + 6x + 9. X 3 3 x Figure 5.42: Find the area of the square. Now that we have the squaring a binomial shortcut, we can simplify the process of finding the area of the outer square by squaring its side. That is: 5.7. SPECIAL PRODUCTS 379 Now we can use the squaring the binomial technique to expand. = x 2 + 2(a;)(3) + (3) 2 = x 2 + 6x + 9 Note that this is the same as the answer found by summing the four parts of the square in Example 7. 380 CHAPTER 5. POLYNOMIAL FUNCTIONS f> t* t* Exercises «•* -*$ •** In Exercises 1-12, use the FOIL shortcut as in Examples 3 and 4 to multiply the given binomials. 1. (5a; + 2)(3x + 4) 7. (6a; - 2) (3a: - 5) 2. (5a; + 2)(4x + 3) 8. (2x - 3) (6a; - 4) 3. (6a; - 3)(5x + 4) 9. (6a; + 4)(3aj + 5) 4. (6a; -2) (4x + 5) 10. (3a; + 2) (4a; + 6) 5. (5a; -6) (3a; -4) 11. (4a; - 5) (6a; + 3) 6. (6a; -4) (3a; -2) 12. (3a; - 5) (2a; + 6) In Exercises 13-20, use the difference of squares shortcut as in Example 5 to multiply the given bino- mials. 13. (10a: - 12) (10a; + 12) 17. (3a; + 10) (3a; - 10) 14. (10a; - 11) (10a; + 11) 18. (12a; + 12) (12a; - 12) 15. (6a; + 9) (6a; - 9) 19. (10a; - 9)(10a; + 9) 16. (9a; + 2) (9a; -2) 20. (4a; - 6) (4a; + 6) In Exercises 21-28, use the squaring a binomial shortcut as in Example 8 to expand the given expression. 21. (2a; + 3) 2 25. (7a: + 2) 2 22. (8a; + 9) 2 26. (4a; + 2) 2 23. (9a; - 8) 2 27. (6a; - 5) 2 24. (4a; - 5) 2 28. (4a: - 3) 2 In Exercises 29-76, use the approrpiate shortcut to multiply the given binomials. 29. (11a: - 2) (Use + 2) 33. (56 + 6c)(36 - 2c) 30. (6a; - 7) (6a; + 7) 34. (3r + 2£)(5r - 3t) 31. {7r-5tf 35. (3u + 5v)(3u - 5v) 32. (11m - 9w) 2 36. (11a + 4c) (11a - 4c) 5.7. SPECIAL PRODUCTS 381 37. (96 3 + 10c 5 )(96 3 - 10c 5 38. (9r 5 + 7£ 2 )(9r 5 -7i 2 ) 39. (9s - At)(9s + At) 40. (12x - 7y)(12x + 7y) 41. (7x - 9y)(7x + 9y) 42. (lOr- llt)(10r+llt) 43. (6a-66)(2a + 36) 44. (6r - 5i)(2r + 3i) 45. (10a;- 10)(10x + 10) 46. (12a;-8)(12a: + 8) 47. (4a + 26) (6a - 36) 48. (36 + 6c)(26-4c) 49. (56-4c)(36 + 2c) 50. (36-2c)(46 + 5c) 51. (46-6c)(66-2c) 52. (Ay - Az)(5y - 3z) 53. (llr 5 + 9£ 2 ) 2 54. (llar 3 + 10;z 5 ) 2 55. (Au - Av)(2u - 6v) 56. (4m — 5w)(5u — 6w) 57. 8r 4 + 7t b ) 2 58. [2x 5 + 5y 2 ) 2 59. [Ar + 3t)(4r-3t) 60. ;3r + 4s)(3r-4s) 61. 5r + 6t) 2 62. [12v + 5w) 2 63. [3x - 4)(2x + 5) 64. ;5a;-6)(4a; + 2) 65. ;66 + 4c)(26 + 3c) 66. '3v + 6w)(2v + Aw) 67. ;11m 2 + 8w 3 )(11u 2 -8w 3 68. ;3u 3 + llw 4 )(3u 3 - llw 4 69. ;4y + 3z) 2 70. 116 + 3c) 2 71. [7u - 2v) 2 72. [Ab - 5c) 2 73. '3v + 2w)(5v + 6w) 74. [by + 3z)(Ay + 2z) 75. to-3)(6:c + 2) 76. ;6x-5)(3a; + 2) For each of the following figure, compute the area of the square using two methods, i) Find the area by summing the areas of its parts (see Example 7 on page 348). ii) Find the area by squaring the side of the square using the squaring a binomial shortcut. 77. 78. X 10 X X 13 13 X 382 CHAPTER 5. POLYNOMIAL FUNCTIONS 79. A square piece of cardboard measures 12 inches on each side. Four squares, each having a side of x inches, are cut and removed from each of the four corners of the square piece of cardboard. The sides are then folded up along the dashed lines to form a box with no top. b) 12 X X X X X X X X / / / V / / 12 Find the volume of the box as a function of x, the measure of the side of each square cut from the four corner of the original piece of cardboard. Multiply to place your answer in standard polynomial form, simplifying your answer as much as possible. Use the resulting polynomial to determine the volume of the box if squares of length 1.25 inches are cut from each corner of the original piece of cardboard. Round your answer to the nearest cubic inch. 80. Consider again the box formed in Exercise 79. a) Find the surface area of the box as a function of x, the measure of the side of each square cut from the four corner of the original piece of cardboard. Multiply to place your answer in standard polynomial form, simplifying your answer as much as possible. b) Use the resulting polynomial to determine the surface area of the box if squares of length 1.25 inches are cut from each corner of the original piece of cardboard. Round your answer to the nearest square inch. £»• s*> £*■ Answers ■** •** ■*$ 1. 15a; 2 + 26a; + 8 3. 30a; 2 + 9a;- 12 5. 15a; 2 - 38a; + 24 7. 18a; 2 - 36a; + 10 9. 18a; 2 + 42a; + 20 11. 24a; 2 - 18a;- 15 13. 100a; 2 - 144 15. 36a; 2 - 81 17. 9a; 2 - 100 19. 100a; 2 - 81 21. 4x 2 + 12a; + 9 23. 81a; 2 - 144a; + 64 5.7. SPECIAL PRODUCTS 383 25. 49a; 2 + 28a; + 4 27. 36a; 2 - 60a; + 25 29. 121a; 2 - 4 31. 49r 2 - 70rt + 2bt 2 33. 156 2 + 86c - 12c 2 35. 9m 2 - 25v 2 37. 816 6 - 100c 10 39. 81s 2 - 16t 2 41. 49a; 2 - 81y 2 43. 12a 2 + 6a6 - 18b 2 45. 100a; 2 - 100 47. 24a 2 - 6b 2 49. 156 2 - 26c - 8c 2 51. 246 2 - 446c + 12c 2 53. 121r 10 + 198r 5 t 2 + 81i 4 55. 8m 2 - 32uv + 2Av 2 57. 64r 8 + 112r 4 t 5 + 49i 10 59. 16r 2 - 9t 2 61. 25r 2 + 60rt + 36t 2 63. 6a; 2 + 7a; - 20 65. 126 2 + 266c + 12c 2 67. 121w 4 - 64w 6 69. 16y 2 + 24yz + 9z 2 71. 49m 2 -28uv + 4v 2 73. 15w 2 + 28vw + 12w 2 75. 30a; 2 - 8a; - 6 77. A = x 2 + 20x + 25 79. a) V(x) = 144a;-48a; 2 +4a; 3 b) V{2) = 128 cubic inches Chapter 6 Factoring The ancient Babylonians left the earliest evidence of the use of quadratic equa- tions on clay tablets dating back to 1800 BC. They understood how the area of a square changes with the length of its side. For example, they knew it was possible to store nine times more bales of hay in a square loft if the side of the loft was tripled in length. However, they did not know how to calculate the length of the side of a square starting from a given area. The word "quadratic" comes from "quadratus," the Latin word for "square." In this chapter, we will learn how to solve certain quadratic equations by factoring polynomials. 385 386 CHAPTER 6. FACTORING You Try It! 6.1 The Greatest Common Factor We begin this section with definitions of factors and divisors. Because 24 = 2 ■ 12, both 2 and 12 are factors of 24. However, note that 2 is also a divisor of 24, because when you divide 24 by 2 you get 12, with a remainder of zero. Similarly, 12 is also a divisor of 24, because when you divide 24 by 12 you get 2, with a remainder of zero. Factors and divisors. Suppose m and n are integers. Then mis a divisor (factor) of n if and only if there exists another integer k so that n = m ■ k. The words divisor and factor are equivalent. They have the same meaning. List the positive divisors of EXAMPLE 1. List the positive divisors (factors) of 24. 18. Solution: First, list all possible ways that we can express 24 as a product of two positive integers: 24 = 1 ■ 24 or 24 = 2 ■ 12 or 24 = 3 ■ 8 or 24 = 4 ■ 6 Answer: 1,2,3,6,9, and 18 Therefore, the positive divisors (factors) of 24 are 1, 2, 3, 4, 6, 8, and 24. □ You Try It! List the positive divisors that EXAMPLE 2. List the positive divisors (factors) that 36 and 48 have in 40 and 60 have in common. common. Solution: First, list all positive divisors (factors) of 36 and 48 separately then box the divisors that are in common. 12 Divisors of 36 are: |T| , |J] , |T| , |T| , |T| , 9, Divisors of 48 are: [T| , [T| , [Y| , [T| , [T| , 8,|~12~|, 16, 24, 48 Answer: 1,2,4,5,10, and 20 Therefore, the common positive divisors (factors) of 36 and 48 are 1, 2, 3, 4, 6, and 12. □ 6. 1 . THE GREATEST COMMON FACTOR 387 Greatest common divisor. The greatest common divisor (factor) of a and b is the largest positive number that divides evenly (no remainder) both a and b. The greatest common divisor of a and 6 is denoted by the symbolism GCD(a, b). We will also use the abbreviation GCF(a, b) to represents the greatest common factor of a and b. Remember, greatest common divisor and greatest common factor have the same meaning. In Example 2, we listed the common positive divisors of 36 and 48. The largest of these common divisors was 12. Hence, the greatest common divisor (factor) of 36 and 48 is 12, written GCD(36, 48) = 12. With smaller numbers, it is usually easy to identify the greatest common divisor (factor). You Try It! EXAMPLE 3. State the greatest common divisor (factor) of each of the State the greatest common following pairs of numbers: (a) 18 and 24, (b) 30 and 40, and (c) 16 and 24. divisor of 36 and 60. Solution: In each case, we must find the largest possible positive integer that divides evenly into both the given numbers. a) The largest positive integer that divides evenly into both 18 and 24 is 6. Thus, GCD(18,24) = 6. b) The largest positive integer that divides evenly into both 30 and 40 is 10. Thus, GCD(30,40) = 10. c) The largest positive integer that divides evenly into both 16 and 24 is 8. Thus, GCD(16,24) = 8. Answer: 12 With larger numbers, it is harder to identify the greatest common divisor (factor). However, prime factorization will save the day! □ EXAMPLE 4. Find the greatest common divisor (factor) of 360 and 756. Solution: Prime factor 360 and 756, writing your answer in exponential form. You Try It! Find the greatest common divisor of 120 and 450. 388 CHAPTER 6. FACTORING 360 756 36 10 9 / \ ($f \) (sf (D ci) fa 3) 4 Thus: 360 = 2 3 • 3 2 ■ 5 756 = 2 2 • 3 3 ■ 7 indexgreatest common divisorlusing prime factorization To find the greatest common divisor (factor), list each factor that appears in common to the highest power that appears in common. In this case, the factors 2 and 3 appear in common, with 2 2 being the high- est power of 2 and 3 2 being the highest power of 3 that appear in common. Therefore, the greatest common divisor of 360 and 756 is: GCD(360,756) = 2 2 -3 2 = 4-9 = 36 Therefore, the greatest common divisor (factor) is GCD(360,756) = 36. Note what happens when we write each of the given numbers as a product of the greatest common factor and a second factor: 360 = 36-10 756 = 36-21 In each case, note how the second second factors (10 and 21) contain no addi- Answer: 30 tional common factors. □ Finding the Greatest Common Factor of Monomials Example 4 reveals the technique used to find the greatest common factor of two or more monomials. 6. 1 . THE GREATEST COMMON FACTOR 389 Finding the GCF of two or more monomials. To find the greatest com- mon factor of two or more monomials, proceed as follows: 1 . Find the greatest common factor (divisor) of the coefficients of the given monomials. Use prime factorization if necessary. 2. List each variable that appears in common in the given monomials. 3. Raise each variable that appears in common to the highest power that appears in common among the given monomials. EXAMPLE 5. Find the greatest common factor of 6x 3 y 3 and 9x 2 y 5 . Solution: To find the GCF of 6x 3 y 3 and 9x 2 y 5 , we note that: 1. The greatest common factor (divisor) of 6 and 9 is 3. 2. The monomials Qx 3 y 3 and 9x 2 y 5 have the variables x and y in common. 3. The highest power of x in common is x 2 . The highest power of y in common is y 3 . Thus, the greatest common factor is GCF(6cc 3 j/ 3 , 9x 2 y 5 ) = 2>x 2 y 3 . Note what happens when we write each of the given monomials as a product of the greatest common factor and a second monomial: 6x 3 y 3 3x 2 y 3 ■ 2x 9 X 2 y 5 = 3 X 2 y 3 ■ 3y Observe that the set of second monomial factors (2a; and 3j/) contain no addi- tional common factors. You Try It! Find the greatest common factor of 16xy 3 and 12x 4 y 2 . Answer: Axy 2 □ EXAMPLE 6. Find the greatest common factor of 12.x 4 , 18a: 3 , and 30a; 2 . Solution: To find the GCF of 12a; 4 , 18a; 3 , and 30a; 2 , we note that: 1. The greatest common factor (divisor) of 12, 18, and 30 is 6. 2. The monomials 12a; 4 , 18a; 3 , and 30a; 2 have the variable x in common. 3. The highest power of x in common is a; 2 . You Try It! Find the greatest common factor of 6y 3 , 15y 2 , and 9j/ 5 . 390 CHAPTER 6. FACTORING Thus, the greatest common factor is GCF(12a; 4 , 18a; 3 , 30a; 2 ) = 6a: 2 . Note what happens when we write each of the given monomials as a product of the greatest common factor and a second monomial: 12a; 4 18a; 3 30a: 2 6a; ■2x 6a; ■ 2>x 6x 2 •5 Answer: 3y 2 Observe that the set of second monomial factors (2a; 2 , 3a;, and 5) contain no additional common factors. □ Factor Out the GCF In Chapter 5, we multiplied a monomial and polynomial by distributing the monomial times each term in the polynomial. 2a;(3a; 2 + 4a; - 7) = 2x ■ 3x 2 + 2x ■ 4x - 2x ■ 7 = 6x 3 + 8a; 2 - 14a; In this section we reverse that multiplication process. We present you with the final product and ask you to bring back the original multiplication problem. In the case 6a; 3 + 8a; 2 — 14a;, the greatest common factor of 6a; 3 , 8a; 2 , and 14a; is 2a;. We then use the distributive property to factor out 2a; from each term of the polynomial. 6a; 3 + 8a; 2 - 14a; = 2x ■ 3a; 2 + 2a; • 4a; - 2x • 7 = 2a;(3a; 2 + 4a; - 7) Factoring. Factoring is "unmultiplying." You are given the product, then asked to find the original multiplication problem. First rule of factoring. If the terms of the given polynomial have a greatest common factor (GCF), then factor out the GCF. Let's look at a few examples that factor out the GCF. You Try It! Factor: 9y 2 - I5y + 12 EXAMPLE 7. Factor: 6a; 2 + 10a; + 14 6. 1 . THE GREATEST COMMON FACTOR 391 Solution: The greatest common factor (GCF) of 6a; 2 , 10a: and 14 is 2. Factor out the GCF. 6a; 2 + 10a; + 14 = 2 ■ 3a; 2 + 2 ■ 5x + 2 ■ 7 = 2(3a; 2 + 5a; + 7) Checking your work. Every time you factor a polynomial, remultiply to check your work. Check: Multiply. Distribute the 2. 2(3a; 2 + 5x + 7) = 2 ■ 3a; 2 + 2 ■ 5x + 2 ■ 7 = 6a: 2 + 10a: + 14 That's the original polynomial, so we factored correctly. Answer: 3(3y 2 — 5y + 4) □ You Try It! EXAMPLE 8. Factor: 12y 5 - 32y 4 + 8y 2 Solution: The greatest common factor (GCF) of I2y 5 , 32j/ 4 and 8y 2 is Ay 2 . Factor out the GCF. 12y 5 - 32y 4 + 8y 2 = 4y 2 • 3y 3 - Ay 2 ■ 8y 2 + Ay 2 ■ 2 = 4y 2 (3j/ 3 -8 2 / 2 + 2) Check: Multiply. Distribute the monomial Ay 2 . Ay 2 {3y 3 - 8y 2 + 2) = 4y 2 • 3y 3 - Ay 2 ■ 8y 2 + Ay 2 ■ 2 = 12y 5 - 32y 4 + 8y 2 That's the original polynomial. We have factored correctly. Factor: 8a; 6 + 20.x 4 - 24a; 3 Answer: 4a; 3 (2x 3 + 5x - 6) □ EXAMPLE 9. Factor: 12a 3 6 + 24a 2 6 2 + 12a6 3 Solution: The greatest common factor (GCF) of 12a 3 6, 24a 2 6 2 and 12a6 3 is 12ab. Factor out the GCF. 12a 3 6 + 24a 2 & 2 + 12a6 3 = 12a6 • a 2 - 12ab ■ 2ab + 12ab ■ b 2 You Try It! Factor: 15s 2 t 4 + 6s 3 t 2 + 9s 2 t 2 Ylabia 2 + 2ab + b 2 392 CHAPTER 6. FACTORING Answer: 3s 2 t 2 (5t 2 + 2s + 3) Check: Multiply. Distribute the monomial Ylab. 12ab(a 2 + 2ab + b 2 ) 12ab ■ a 2 - 12ab ■ 2ab + \2ab ■ b 2 ,2,2 = YlaTb + 24a z b A + \2ab a That's the original polynomial. We have factored correctly. □ You Try It! Factor: 18p 5 q 4 - 30pV + 42pV Answer: 6p 3 q i (3p 2 -5pq + 7q 2 ) Speeding Things Up a Bit Eventually, after showing your work on a number of examples such as those in Examples 7, 8, and 9, you'll need to learn how to perform the process mentally. EXAMPLE 10. Factor each of the following polynomials: (a) 24a; + 32, (b) 5a; 3 - 10a; 2 - 10a;, and (c) 2a; 4 y + 2a; 3 y 2 - 6x 2 y 3 . Solution: In each case, factor out the greatest common factor (GCF): a) The GCF of 24a; and 32 is 8. Thus, 24a; + 32 = 8(3a: + 4) b) The GCF of 5a; 3 , 10x 2 , and 10a; is 5a;. Thus: 5a; 3 - 10a; 2 - 10a; = 5x(a; 2 - 2x - 2) c) The GCF of 2a; 4 y, 2a; 3 y 2 , and 6x 2 y 3 is 2a; 2 y. Thus: 2x i y + 2x 3 y 2 Qx 2 y 3 2x y(x + xy — 3y ) As you speed things up by mentally factoring out the GCF, it is even more important that you check your results. The check can also be done mentally. For example, in checking the third result, mentally distribute 2x 2 y times each term of x 2 + xy — 3y 2 . Multiplying 2x 2 y times the first term a; 2 produces 2x y, the first term in the original polynomial. 1- „2„,3 2a;^ (x z +xy- 3y z ) = 2x*y + 2x 3 y z - Qx z y I t Continue in this manner, mentally checking the product of 2a; 2 y with each term of x 2 xy 3y 2 , making sure that each result agrees with the corresponding term of the original polynomial. □ Remember that the distributive property allows us to pull the GCF out in front of the expression or to pull it out in back. In symbols: ab + ac = a(b + c) ba + ca = (b + c)a 6. 1 . THE GREATEST COMMON FACTOR 393 EXAMPLE 11. Factor: 2x{3x + 2) + 5(3x + 2) Solution: In this case, the greatest common factor (GCF) is 3x + 2. 2x(3x + 2) + 5(3x + 2) = 2x ■ (3x + 2) + 5 • (3x + 2) = (2x + 5)(3x + 2) Because of the commutative property of multiplication, it is equally valid to pull the GCF out in front. 2x(3x + 2) + 5(3x + 2) = (3x + 2) • 2x + (3x + 2) • 5 = (3x + 2)(2x + 5) Note that the order of factors differs from the first solution, but because of the commutative property of multiplication, the order does not matter. The answers are the same. You Try It! Factor: 3x 2 (4x- 7) + 8(4x - 7) Answer: (3a; 2 + 8) (4a; - 7) □ EXAMPLE 12. Factor: 15a(o + b) - 12 (a + b) Solution: In this case, the greatest common factor (GCF) is 3(a + b). 15a(a + b) - 12(o + i 3(a + 6)-5a-3(a + 6) -4 3(a + 6)(5a-4) You Try It! Factor: 2Am(m - 2n) + 20(m - 2n) Alternate solution: It is possible that you might fail to notice that 15 and 12 are divisible by 3, factoring out only the common factor a + b. 15a(a + b) - 12(a + i 15a- (a + 6)- 12- (a + b) (15a- 12) (a + 6) However, you now need to notice that you can continue, factoring out 3 from both 15a and 12. = 3(5a-4)(a + 6) Note that the order of factors differs from the first solution, but because of the commutative property of multiplication, the order does not matter. The answers are the same. Answer: 4(6m + 5)(m — 2n) □ 394 CHAPTER 6. FACTORING You Try It! Factor by grouping: x 2 - 6x + 2x - 12 Answer: (x + 2)(x — 6) Factoring by Grouping The final factoring skill in this section involves four-term expressions. The technique for factoring a four-term expression is called factoring by grouping. EXAMPLE 13. Factor by grouping: x 2 + 8x + 3x + 24 Solution: We "group" the first and second terms, noting that we can factor an x out of both of these terms. Then we "group" the third and fourth terms, noting that we can factor 3 out of both of these terms. x 2 + 8x + 3x + 24 = x (x + 8) + 3 (x + 8) L J L J Now we can factor x + 8 out of both of these terms. (x + 3)(x + . D You Try It! Factor by grouping: x 2 - 5x - Ax + 20 Answer: (a; — 4)(.t — 5) Let's try a grouping that contains some negative signs. EXAMPLE 14. Factor by grouping: x 2 + Ax - 7x - 28 Solution: We "group" the first and second terms, noting that we can factor x out of both of these terms. Then we "group" the third and fourth terms, then try to factor a 7 out of both these terms. x 2 + Ax - 7x - 28 = x (x + A) + 7 (-x - A) { I t i This does not lead to a common factor. Let's try again, this time factoring a — 7 out of the third and fourth terms. x 2 + Ax - 7x - 28 = x (x + 4) - 7 (x + A) t I { i That worked! We now factor out a common factor x + A. = (x-7){x + A) D 6. 1 . THE GREATEST COMMON FACTOR 395 Let's increase the size of the numbers a bit. EXAMPLE 15. Factor by grouping: 6a; 2 - 8a; + 9x - 12 Solution: Note that we can factor 2x out of the first two terms and 3 out of the second two terms. 6a; 2 - 8a; + 9a; - 12 = 2a; (3a; - 4) + 3 (3a; - 4) t i t f Now we have a common factor 3a; — 4 which we can factor out. You Try It! Factor by grouping: 15a; 2 + 9a; + 10a; + 6 (2a; + 3)(3a;-4) Answer: (3a; + 2) (5a; + 3) As the numbers get larger and larger, you need to factor out the GCF from each grouping. If not, you won't get a common factor to finish the factoring. □ EXAMPLE 16. Factor by grouping: 24a; 2 - 32a; - 45a; + 60 Solution: Suppose that we factor 8a; out of the first two terms and —5 out of the second two terms. You Try It! Factor by grouping: 36a; 2 - 84a; + 15a; - 35 24a; 2 t_ 32a; - 45a; + 60 = 8a; (3a; - 4) - 5 (9a; - 12) t t I That did not work, as we don't have a common factor to complete the factoring process. However, note that we can still factor out a 3 from 9a; — 12. As we've already factored out a 5, and now we see can factor out an additional 3, this means that we should have factored out 3 times 5, or 15, to begin with. Let's start again, only this time we'll factor 15 out of the second two terms. 24a; 2 t_ 32a; - 45a; _f L_ 60 8x (3a; - 4) - 15 (3a; - 4) Beautiful! We can now factor out 3a; — 4. 3a;- 15)(3a;-4) Answer: (12a; + 5) (3a; - 7) □ 396 CHAPTER 6. FACTORING f> t* t* Exercises ■** •** ■** In Exercises 1-6, list all positive divisors of the given number, in order, from smallest to largest. 1. 42 4. 85 2. 60 5. 51 3. 44 6. 63 In Exercises 7-12, list all common positive divisors of the given numbers, in order, from smallest to largest. 7. 36 and 42 10. 96 and 78 8. 54 and 30 11. 8 and 76 9. 78 and 54 12. 99 and 27 In Exercises 13-18, state the greatest common divisor of the given numbers. 13. 76 and 8 16. 64 and 76 14. 84 and 60 17. 24 and 28 15. 32 and 36 18. 63 and 27 In Exercises 19-24, use prime factorization to help calculate the greatest common divisor of the given numbers. 19. 600 and 1080 22. 540 and 150 20. 150 and 120 23. 600 and 450 21. 1800 and 2250 24. 4500 and 1800 In Exercises 25-36, find the greatest common factor of the given expressions. 25. 166 4 and 566 9 28. 24w 3 and 30w 8 26. 28,s 2 and 36s 4 29. 56ccV and 16a; V 27. 35^ 2 and 49z 7 30. 356 5 c 3 and 636 4 c 4 6.1. THE GREATEST COMMON FACTOR 397 31. 24s 4 i 5 and 16s 3 i 6 34. 8r 7 , 24r 6 , and 12r 5 32. 10v 4 w 3 and 8v 3 w 4 35. 9a 6 , 6a 5 , and 15a 4 33. I8y 7 , 45y 6 , and 27y 5 36. 15a 5 , 24a 4 , and 24a 3 In Exercises 37-52, factor out the GCF from each of the given expressions. 37. 25a 2 + 10a +20 45. 35s 7 + 49s 6 + 63s 5 38. 40c 2 + 15c + 40 46. 35s 7 + 56s 6 + 56s 5 39. 35s 2 + 25s + 45 47. 146 7 + 356 6 + 566 5 40. 456 2 + 206 + 35 48. 45a; 5 + 81a; 4 + 45a; 3 41. 16c 3 + 32c 2 + 36c 49. 54y 5 z 3 + 30yV + 36y 3 z 5 42. 126 3 + 126 2 + 186 50. 42a; 4 ?/ 2 + 42x 3 y 3 + 54x 2 y 4 43. 42s 3 + 24s 2 + 18s 51. 45s 4 * 3 + 40s 3 t 4 + 15s 2 i 5 44. 36y 3 + 81y 2 + 36y 52. 20v 6 w 3 + 36v 5 w 4 + 28v 4 w 5 In Exercises 53-60, factor out the GCF from each of the given expressions. 53. 7w{2w - 3) - 8(2w - 3) 57. 48a(2a + 5) - 42(2a + 5) 54. 5s(8s - 1) + 4(8s - 1) 58. 40«(7w - 4) + 72(7w - 4) 55. 9r(5r - 1) + 8(5r - 1) 59. 56a(2a - 1) - 21(2a - 1) 56. 5c(4c - 7) + 2(4c - 7) 60. 48r(5r + 3) - 40(5r + 3) In Exercises 61-68, factor by grouping. Do not simplify the expression before factoring. 61. x 2 + 2x - 9x- 18 65. x 2 - 6x - 3x + 18 62. x 2 + 6a-- 9a;- 54 66. a; 2 - 3a; - 9a; + 27 63. a; 2 + 3a; + 6a; + 18 67. a; 2 - 9a; + 3a; - 27 64. a; 2 + 8a; + 7a; + 56 68. a; 2 - 2a; + 7a; - 14 398 CHAPTER 6. FACTORING In Exercises 69-76, factor by grouping. Do not simplify the expression before factoring. 69. 8a; 2 + 3a; - 56a; - 21 73. 6a; 2 - 7s - 48a; + 56 70. 4a; 2 + 9a; - 32a; - 72 74. 8a; 2 - 7a; - 72a; + 63 71. 9a; 2 + 36a; - 5a; - 20 75. 2a; 2 + 12s + 7a; + 42 72. 7a; 2 + 14s - 8s - 16 76. 7a; 2 + 28a; + 9s + 36 £»• s*> s*> Answers >** ~*$ •** 1. {1, 2, 3, 6, 7, 14, 21, 42} 39. 5(7s 2 + 5s + 9) 3. {1, 2, 4, 11, 22, 44} 41. 4c(4c 2 + 8c + 9) 5. {1,3,17,51} 43. 6s(7s 2 + 4s + 3) 7.(1,2,3,6} 45. 7s 5 (5s 2 + 7s + 9) 9.(1,2,3,6} 47. 76 5 (26 2 + 56 + 8) 11. {1,2,4} 49. 6y 3 z 3 {9y 2 + 5yz + 6z 2 ) 13.4 51.5s 2 t 3 (9s 2 + 8st + 3t 2 ) 15.4 53. (7w-8)(2w-3) 17. 4 55. (9r + 8)(5r- 1) 19.120 57. 6(2a + 5)(8a- 7) 21.450 59. 7(2a- l)(8a-3) 23. 150 61. (s-9)(s + 2) 25. 86 4 63. (s + 6)(a; + 3) 27. 7z 2 65. (s-3)(s-6) 29. 8x 2 y 4 67. (s + 3)(s-9) 31. 8s 3 t 5 69. (s- 7)(8a; + 3) 33. 9y 5 71. (9s-5)(s + 4) 35. 3a 4 73. (s-8)(6s- 7) 37. 5(5a 2 + 2a + 4) 75. (2s + 7)(s + 6) 6.2. SOLVING NONLINEAR EQUATIONS 399 6.2 Solving Nonlinear Equations We begin by introducing a property that will be used extensively in this and future sections. The zero product property. If the product of two or more numbers equals zero, then at least one of the numbers must equal zero. That is, if ab = 0, then a = or 6 = 0. Let's use the zero product property to solve a few equations. You Try It! EXAMPLE 1. Solve for a;: (x + 3) (a; - 5) = Solve for a;: Solution: The product of two factors equals zero. (x + 3)(x-5) =0 Hence, at least one of the factors must equal zero. Using the zero product property, set each factor equal to zero, then solve the resulting equations for x. £ + 3 = or x — 5 = x = —3 x = 5 Hence, the solutions are x = — 3 and x = 5 Check: Check that each solution satisfies the original equation. Substitute —3 for x: Substitute 5 for x: (x + 3)(x-5) = (x + 3)(x-5)=0 (-3 + 3)(-3-5) = (5 + 3)(5-5) = (0)(-8) = (8)(0) = 0=0 0=0 Because each check produces a true statement, both x = — 3 and x = 5 are solutions of (x + 3)(x — 5) = 0. Answer: 7, 2 The zero product property also works equally well if more than two factors are present. For example, if abc = 0, then either a = or b = or c = 0. Let's use this idea in the next example. □ 400 CHAPTER 6. FACTORING You Try It! Solve for x: Qx(x + 4) (5a; + 1) = Answer: 0, -1/5 EXAMPLE 2. Solve for x: x(2x + 9) (3a; - 5) = Solution: The product of three factors equals zero. a;(2a; + 9)(3a;-5) = Using the zero product property, set each factor equal to zero, then solve the resulting equations for x. x = or 2x + 9 = or 3a; - 5 = 2a; = -9 3a; = 5 9 x = — 2 5 x = — 3 Hence, the solutions are x = 0, x = —9/2, and x = 5/3. We encourage the reader to check the solution. □ Linear versus Nonlinear All of the equations solved in previous chapters were examples of what are called linear equations. If the highest power of the variable we are solving for is one, then the graphs involved are lines. Hence the term, linear equation. However, if the power on the variable we are solving for exceeds one, then the graphs involved are curves. Hence the term, nonlinear equation. In this chapter we will learn how to solve nonlinear equations involving polynomials. However, let's first make sure we can recognize the difference between a linear and a nonlinear equation. Linear versus Nonlinear. Use the following conditions to determine if an equation is linear or nonlinear. 1. If the highest power of the variable we are solving for is one, then the equation is linear. 2. If the highest power of the variable we are solving for is larger than one, then the equation is nonlinear. You Try It! Classify the following equation as linear or nonlinear: 2x = x 3 — EXAMPLE 3. If the instruction is "solve for x, equations as linear or nonlinear. classify each of the following 6.2. SOLVING NONLINEAR EQUATIONS 401 a) 3x 7x b) x 2 8x Solution: Because the instruction is "solve for a;," to determine whether the equation is linear or nonlinear, we identify the largest power of x present in the equation. a) The highest power of x present in the equation 3x — 5 Hence, this equation is linear. 7x is one. b) The equation x 2 = 8x contains a power of x higher than one (it contains an x 2 ). Hence, this equation is nonlinear. Answer: nonlinear Now that we can classify equations as either linear or nonlinear, let's in- troduce strategies for solving each type, the first of which should already be familiar. Strategy for solving a linear equation. If an equation is linear, start the solution process by moving all terms containing the variable you are solving for to one side of the equation, then move all terms that do not contain the variable you are solving for to the other side of the equation. □ You Try It! EXAMPLE 4. Solve for 3a; 4 -7a; Solution: Because the instruction is "solve for x" and we note that the largest power of x present is one, the equation 3a: — 5 = 4 — 7x is linear. Hence, the strategy is to move all terms containing x to one side of the equation, then move all the remaining terms to the other side of the equation. Solve for x: 2x + 4 = 6a; 3a; - 5 = 4 - 7x 3a; - 5 + 7x = 4 3a; + 7x = 4 + 5 Original equation. Add 7x to both sides. Add 5 to both sides. Note how we have succeeded in moving all terms containing x to one side of the equation and all terms that do not contain x to the other side of the equation. 10a; = 9 Simplify both sides. 9 x = — Divide both sides by 10. 10 Hence, the solution of 3a; — 5 = 4 — 7x is x = 9/10. Readers are encouraged to check this solution. Answer: 1/4 □ 402 CHAPTER 6. FACTORING You Try It! Solve for x: x 2 -5x The situation is much different when the equation is nonlinear. Strategy for solving a nonlinear equation. If an equation is nonlinear, first move everything to one side of the equation, making one side of the equa- tion equal to zero. Continue the solution process by factoring and applying the zero product property. EXAMPLE 5. Solve for x: 8x Solution: Because the instruction is "solve for a;," and the highest power of x is larger than one, the equation x 2 = 8x is nonlinear. Hence, the strategy requires that we move all terms to one side of the equation, making one side zero. x z = 8x 8a; = Original equation. Subtract 8a; from both sides. Note how we have succeeded in moving all terms to one side of the equation, making one side equal to zero. To finish the solution, we factor out the GCF on the left-hand side. x(a;-8) = Factor out the GCF. Note that we now have a product of two factors that equals zero. By the zero product property, either the first factor is zero or the second factor is zero. Hence, the solutions are x = and x = 8. Check: Check that each solution satisfies the original equation. Substitute for x: Subtitute 8 for x: Answer: 0, x 2 =8x (0) 2 = 8(0) = x (8) 2 64 64 Note that both results are true statements, guaranteeing that both x = and x = 8 are solutions of x 2 = 8x. □ 6.2. SOLVING NONLINEAR EQUATIONS 403 Warning! The following is incorrect! Consider what would happen if we divided both sides of the equation x 2 = 8x in Example 5 by a;: x z = 8x x 2 _ 8x X X Note that we have lost the second answer found in Example 5, x = 0. This example demonstrates that you should never divide by the variable you are solving for! If you do, and cancellation occurs, you will lose answers. Let's try solving a nonlinear equation that requires factoring by grouping. You Try It! EXAMPLE 6. Solve for x: 6x 2 + 9x - 8x - 12 = Solve for x: 5a; 2 — 20a; — 4a; + 16 = Solution: Because we are solving for x and there is a power of x larger than one, this equation is nonlinear. Hence, the first step is to move everything to one side of the equation, making one side equal to zero. Well that's already done, so let's factor the left-hand side by grouping. Note that we can factor 3a; out of the first two terms and —4 out of the second two terms. J \ | \ 6x 2 + 9x - 8x - 12 = 3a; (2a; + 3) - 4 (2a; + 3) = Factor out the common factor 2a; + 3. (3a;-4)(2a; + 3) = We now have a product of two factors that equals zero. Use the zero product property to write: 3a; - 4 = or 2a; + 3 = 3a; = 4 2a; = -3 4 X= 3 3 X = ~2 Hence, the solutions are x = 4/3 and x = —3/2. Check. Let's use the graphing calculator to check the solution x = 4/3. First, store the solution 4/3 in the variable X using the following keystrokes (see the first image in Figure 6.1. 404 CHAPTER 6. FACTORING \T\ 3 EE X,T,9,n ENTER Next, enter the left-hand side of the equation 6a; 2 + 9a; — 8a;— 12 = using the following keystrokes. Note that the result in the second image in Figure 6.1 indicates that the expression 6a; 2 + 9a; — 8a; — 12 equals zero when x = 4/3. X,T,0,n X,T,0,« m mm ENTER Answer: 4/5, 4 4/3+X 1 . 333333333 6*X A 2+9*X-8*X-12 Figure 6.1: Checking the solution x = 4/3. Therefore, the solution x = 4/3 checks. Readers are encouraged to use their graphing calculators to check the second solution, x = —3/2. □ Using the Graphing Calculator In this section we will employ two different calculator routines to find the solution of a nonlinear equation. Before picking up the calculator, let's first use an algebraic method to solve the equation a; 2 = —5a;. The equation is nonlinear, so the first step is to move everything to one side of the equation, making one side equal to zero. x = - x 2 + 5a; = a;(a; + 5) = 5 a; Nonlinear. Make one side zero. Add 5a; to both sides. Factor out the GCF. Use the zero product property, setting each factor equal to zero, then solving the resulting equations for x. or x + 5 = x = —5 Hence, the solutions are x = and x 6.2. SOLVING NONLINEAR EQUATIONS 405 We'll now use the calculator to find the solutions of -5x. The first technique employs the 5:intersect routine on the calculator's CALC menu. EXAMPLE 7. Use the 5:intersect utility on the graphing calculator to solve the equation x 2 = — 5x for x. Solution: Load the left-hand side of x 2 = —5x in Yl and the right-hand side in Y2 (see Figure 6.2). Selecting 6:ZStandard from the ZOOM menu produces the graphs shown in the image on the right in Figure 6.2. You Try It! Use the 5: intersect utility on the graphing calculator to solve the equation x 2 = Ax for x. Plotl PlotE Plots E / WiBX A 2 ; / ^.VeB-5+X :/ •J. We = \Vs= \V?= Figure 6.2: Sketch the graphs of each side of the equation x 2 -hx. Note that the graph of y = x 2 is a parabola that opens upward, with vertex (turning point) at the origin. This graph reveals why the equation x 2 = —5x is called a nonlinear equation (not all the graphs involved are lines). Next, the graph of y = —5x is a line with slope —5 and y-intercept at the origin. The two graphs obviously intersect at the origin, but it also appears that there may be another point of intersection that is off the screen. Let's increase Ymax in an attempt to reveal the second point of intersection. After some experimentation, the settings shown in the first image in Figure 6.3 reveal both points of intersection. Pushing the GRAPH button produces the image on the right in Figure 6.3. WINDOW Xnin=-10 Xnax=10 Xscl=l Ymin=-20 Vnax=5@ Yscl=10 Figure 6.3: Adjust the WINDOW parameters to reveal both points of inter- section. To find the solutions of the equation x 2 of the points where the graphs of y = -5x, we must find the coordinates and y -5x intersect. The x- coordinate of each point of intersection will be a solution of the equation x 2 —5x. 406 CHAPTER 6. FACTORING • Start by selecting 5:intersect from the CALC menu. When prompted for the "First curve?" , press ENTER. When prompted for the "Second curve?", press ENTER. When prompted for a "Guess," press ENTER. The result is the point (0, 0) shown in the image on the left in Figure 6.4. • Repeat the process a second time. Select 5:intersect from the CALC menu. When prompted for the "First curve?" , press ENTER. When prompted for the "Second curve?" , press ENTER. When prompted for a "Guess," use the left-arrow key to move the cursor closer to the leftmost point of intersection, then press ENTER. The result is the point (—5, 25) shown in the image on the right in Figure 6.4. \ Intersection^ K=0 / Y=0 ■V Intersection' 7=16 ■J Intersection K=0 V=0\ ■J Intersection K=-£ Y=££ Figure 6.4: Use the 5:intersect utility to find the points of intersection. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with x and y, respectively (see Figure 6.5). • Place your WINDOW parameters at the end of each axis (see Figure 6.5). • Label each graph with its equation (see Figure 6.5). • Drop dashed vertical lines through each point of intersection. Shade and label the x- values of the points where the dashed vertical line crosses the x-axis. These are the solutions of the equation x 2 = — 5x (see Figure 6.5). Hence, the solutions of x 2 = — 5x are x = —5 and x = 0. Note now these match the solutions found using the algebraic technique. □ Before demonstrating a second graphing calculator technique for solving nonlinear equations, let's take a moment to recall the definition of a zero of a function, which was first presented in Chapter 5, Section 3. Zeros and ^-intercepts. The points where the graph of / crosses the x-axis are called the x-intercepts of the graph of /. The x-value of each x-intercept is called a zero of the function /. 6.2. SOLVING NONLINEAR EQUATIONS 407 y = — 5x Figure 6.5: Reporting your graphical solution on your homework. We'll now employ the 2:zero utility from the CALC menu to find the solutions of the equation x 2 = — 5x. You Try It! EXAMPLE 8 the equation x 2 Solution: First, make one side of the equation equal to zero Use the 2:zero utility on the graphing calculator to solve — 5x for x. Use the 2:zero utility on the graphing calculator to solve the equation x 2 = Ax for x. x 5x -5x Make one side zero. Add 5a; to both sides. To determine the values of x that make x 2 + 5x = 0, we must locate the points where the graph of f(x) = x 2 + 5x crosses the a>axis. These points are the ai-intercepts of the graph of / and the a;- values of these points are the zeros of the function /. Load the function f(x) = x 2 + 5x in Yl, then select 6:ZStandard to produce the image in Figure 6.6. Note that the graph of / has two ai-intercepts, and the a;- values of each of these points are the zeros of the function /. Plotl PlotE Plots WiBX A 2+5*X -■■Vi = Wj = We = J It's often easier to find the so- lutions of a nonlinear equa- tion by making one side zero and identifying where the graph of the resulting func- tion crosses the a>axis. Figure 6.6: Sketch the graph of p(x) hx. Select 2:zero from the CALC menu (see Figure 6.7). 408 CHAPTER 6. FACTORING • The calculator responds by asking for a "Left Bound?" Use the left-arrow key to move the cursor so that it lies to the left of the ^-intercept near (—5, 0) (see the second image in Figure 6.7), then press the ENTER key. • The calculator responds by asking for a "Right Bound?" Move the cursor so that is slightly to the right of the x-intercept near (—5, 0) (see the third image Figure 6.7), then press the ENTER key. lilla jBualue Z: zero 3: rdniFiUFi 4: maximun 5: inter-sect 6:dy^dx 7:^fCx>dx 1 Left Bound? K= "£.319149 Y=l.fi97fi007 Y1=!T£+E*H RiSht Bound? K—H.SHOHSl V="1.H93BH9 Figure 6.7: Setting the left and right bounds when using the 2:zero utility to find the ^-intercepts of the graph of f{x) = x 2 + 5x. • The calculator responds by asking for a "Guess?" Note the two triangular marks near the top of the viewing window in the first image in Figure 6.8 that mark the left- and right-bounds. As long as you place the cursor so that the rvalue of the cursor location lies between these two marks, you've made a valid guess. Because the cursor already lies between these two marks, we usually leave it where it is and press the ENTER key. Y1=!T£+E*H 4 Gut::? K=-4.fiH0B£i V="1.H93BH9 t K=-£ V=0 Figure 6.8: Setting the right bound and making a guess. After making your guess and pressing the ENTER key, the calculator pro- ceeds to find an approximation of the ^-intercept that lies between the left- and right-bounds previously marked (see the second image in Figure 6.8). Hence, this ^-intercept is (—5, 0), making —5 a zero of f{x) = x 2 + 5x and a solution of the equation x 2 + 5x = 0. We'll leave it to our readers to repeat the 2:zero process to find the second zero at the origin. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. 6.2. SOLVING NONLINEAR EQUATIONS 409 • Label the horizontal and vertical axes with x and y, respectively (see Figure 6.9). • Place your WINDOW parameters at the end of each axis (see Figure 6.9). • Label each graph with its equation (see Figure 6.9). • Drop dashed vertical lines through each ^-intercept. Shade and label the a:- values of each x-intercept. These are the solutions of the equation -5x (see Figure 6.9). Answer: Z*K* K=0 V=0 Figure 6.9: Reporting your graphical solution on your homework. Hence, the solutions of x 2 = —5a; are x = —5 and x = 0. Note how nicely this agrees with the solutions found using the algebraic technique and the solutions found using the 5:intersect utility in Example 7. ...\l Z*K0 V=0 □ 410 CHAPTER 6. FACTORING *».*».**. Exercises ■** •** * In Exercises 1-8, solve the given equation for x. 1. (9a; + 2) (8a; + 3) =0 5. -9a;(9a; + 4) = 2. (2x-5)(7x-4) = 6. 4a;(3a; - 6) = 3. x(4x + 7)(9» - 8) = 7. (a; + 1)(» + 6) = 4. ar(9a;-8)(3a; + l) = 8. (a? - 4)(x - 1) = In Exercises 9-18, given that you are solving for x, state whether the given equation is linear or nonlinear. Do not solve the equation. 9. x 2 + 7x = 9x + 63 14. 4x 2 = -7x 10. x 2 + 9x = ix + 36 15. 3x 2 + 8x = -9 11. 6a; - 2 = 5a; - 8 16. 5a; 2 - 2a; = -9 12. -5a; + 5 = -6a; - 7 17. -3a; + 6 = -9 13. 7a; 2 = -2a; 18. 8a; - 5 = 3 In Exercises 19-34, solve each of the given equations for x. 19. 3a; + 8 = 9 27. 9a; + 2 = 7 20. 3a; + 4 = 2 28. 3a; + 2 = 6 21. 9a; 2 = -x 29. 9a; 2 = 6a; 22. 6a; 2 = 7a; 30. 6a; 2 = -14a; 23. 3a; + 9 = 8x + 7 31. 7a; 2 = -4x 24. 8a; + 5 = 6a; + 4 32. 7a; 2 = -9a; 25. 8a; 2 = -2a; 33. 7a; + 2 = 4a; + 7 26. 8a; 2 = 18x 34. 4x + 3 = 2x + 8 6.2. SOLVING NONLINEAR EQUATIONS 411 In Exercises 35-50, factor by grouping to solve each of the given equations for x. 35. 63a; 2 + 56a; + 54a; + 48 = 43. x 2 + 6x - 11a; - 66 = 36. 27a; 2 + 36a; + 6a; + 8 = 44. a; 2 + 6a; - 2x - 12 = 37. 16a; 2 - 18a; + 40a; - 45 = 45. 15a; 2 - 24a; + 35a; - 56 = 38. 42a; 2 - 35a; + 54a; - 45 = 46. 12a; 2 - 10a; + 54a; - 45 = 39. 45a; 2 + 18a; + 20a; + 8 = 47. a; 2 + 2a; + 9a; + 18 = 40. 18a; 2 + 21a; + 30a; + 35 = 48. a; 2 + 8a; + 4ai + 32 = 41. a; 2 + 10a; + 4a; + 40 = 49. a; 2 + 4a; - 8a; - 32 = 42. a; 2 + 11a; + 10a; + 110 = 50. a; 2 + 8a; - 5a; - 40 = In Exercises 51-54, perform each of the following tasks: i) Use a strictly algebraic technique to solve the given equation. ii) Use the 5:intersect utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown in Example 7. 51. a; 2 = -Ax 53. a; 2 = 5a; 52. a; 2 = 6a; 54. a; 2 = -6a; In Exercises 55-58, perform each of the following tasks: i) Use a strictly algebraic technique to solve the given equation. ii) Use the 2:zero utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown in Example 8. 55. a; 2 + 7a; = 57. a; 2 - 3a; = 56. a; 2 - 8a; = 58. a; 2 + 2a; = £*• £*< £*■ Answers •*$ •** **$ 2 3 4 1. x = — , — 5. x = 0, — 9' 8 '9 7 8 3. x = 0, --, - ' 4' 9 7. x = -1,-6 9. Nonlinear 11. Linear 13. Nonlinear 15. Nonlinear 17. Linear 19. 1 3 21. 1 x = Q, — 9 23. 2 5 25. 1 x = 0,-- 27. 5 9 29. 2 31. 4 * = 0, -- 33. 5 3 412 CHAPTER 6. FACTORING 68 35. x = --, -- 7' 9 59 37. x= --, - 2' 8 4 2 39. x = --, -- 9' 5 41. x = -4, -10 43. x= 11, -6 7 8 45. x = — , - 3' 5 47. x= -9, -2 49. x = 8, -4 51. x= -4,0 53. x = 0,5 55. x= -7,0 57. x = 0,3 6.3. FACTORING AX 2 + BX + C WHEN A = 1 413 6.3 Factoring ax 2 + bx + c when a = 1 In this section we concentrate on learning how to factor trinomials having the form ax 2 + bx + c when a = 1. The first task is to make sure that everyone can properly identify the coefficients a, b, and c. You Try It! EXAMPLE 1. Compare x 2 — 8x — 9 with the form ax 2 + bx + c and identify Compare 2a; 2 + 5a; — 3 with the coefficients a, 6, and c. Solution: Align the trinomial x 2 — 8x — 9 with the standard form ax 2 + bx + c, then compare coefficients. Note that the understood coefficient of x 2 is 1. the form ax 2 + bx + c and identify the coefficients a, b, and c. ax bx+c lx z - 8a;-9 We see that a = 1, b = — 8, and c = — 9. Because the leading coefficient is 1, this is the type of trinomial that we will learn how to factor in this section. Answer: a = 2,b 5, c □ EXAMPLE 2. Compare -40 + 6x 2 identify the coefficients a, b, and c. x with the form ax 2 + bx + c and Solution: First, arrange —40 + 6a; 2 — x in descending powers of x, then align it with the standard form -bx + c and compare coefficients. Note that the understood coefficient of a; is — 1. You Try It! Compare 3a; + 9 — 7a; 2 with the form ax 2 + bx + c and identify the coefficients a, 6, and c. 6a; 2 bx + c la; -40 We see that a = 6, b = — 1, and c = —40. Because the leading coefficient is 6, we will have to wait until Factoring aa; 2 + bx + c when a/ 1 on page 427 before learning how to factor this trinomial. Answer: a = —7, 3, c = 9 □ In this section, the leading coefficient must equal 1. Our work in this section will focus only on trinomials of the form x 2 + bx + c, that is, the form ax bx + c where a = 1. The ac-Method We are now going to introduce a technique called the ac-method (or ac-test) for factoring trinomials of the form aa; 2 +bx + c when a = 1. In the upcoming 414 CHAPTER 6. FACTORING section Factoring ax 2 + bx + c when a^l on page 427, we will see that this method can also be employed when o / 1, with one minor exception. But for the remainder of this section, we focus strictly on trinomials whose leading coefficient is 1. Let's begin by finding the following product: (x + 12) (a; — 4) = x(x — 4) + 12(cc — 4) Apply the distributive property. Distribut Simplify. X — 4a: + 12 x — 48 Distribute again. ■ 8a; - 48 Now, can we reverse the process? That is, can we start with x 2 + 8x — 48 and place it in its original factored form (x + 12) (x — 4) ? The answer is yes, if we apply the following procedure. The ac-method. Compare the given polynomial with the standard form ax 2 + bx + c, determine the coefficients a, b, and c, then proceed as follows: 1. Multiply the coefficients a and c and determine their product ac. List all the integer pairs whose product equals ac. 2. Circle the pair in the list produced in step 1 whose sum equals b, the coef- ficient of the middle term of ax + c. 3. Replace the middle term bx with a sum of like terms using the circled pair from step 2. 4. Factor by grouping. 5. Check the result using the FOIL shortcut. You Try It! Factor: x 2 + 11a; + 28 Let's follow the steps of the ac-method to factor x 2 + 8x — 48. EXAMPLE 3. Factor: 8a; - 48. Solution: Compare x 2 + 8a; — 48 with ax 2 + bx +c and identify a = 1, b = 8, and c = —48. Note that the leading coefficient is a = 1. Calculate ac. Note that ac = (1)(— 48), so ac = —48. List all integer pairs whose product is ac = —48. 1. -48 -1,48 2. -24 -2,24 3. -16 -3,16 4- -12 -4,12 6. -8 -6,8 6.3. FACTORING AX 2 + BX + C WHEN A = 1 415 Circle the ordered pair whose sum is b -48 -24 -16 -12 -1,48 -2,24 -3,16 -4,12 -6,8 x 2 +8x - -48 = x 2 -ix + 12a; -48 2 + 8x- - 48 = x(x - 4) + 12(sc - 4) = (a: + 12)(a:-4) Replace the middle term 8x with a sum of like terms using the circled pair whose sum is 8. Factor by grouping. Use the FOIL shortcut to mentally check your answer. To determine the prod- uct (x + 12) (a; — 4), use these steps: • Multiply the terms in the "First" positions: x 2 . • Multiply the terms in the "Outer" and "Inner" positions and combine the results mentally: —Ax + Ylx = 8x. • Multiply the terms in the "Last" positions: —48. That is: (a; + 12)(a:-4) F ,,2 o Ax I Ylx L 48 Combining like terms, (x + 12) (x — A) 8a; — 48, which is the original trinomial, so our solution checks. Note that if you combine the "Outer" and "Inner" products mentally, the check goes even faster. Answer: (x + A)(x + 7) □ Some readers might ask "Is it a coincidence that the circled pair —A, 12 seemed to 'drop in place' in the resulting factoriztion (x + 12)(x — 4)?" Before we answer that question, let's try another example. You Try It! EXAMPLE 4. Factor: 9x - 36. Factor: 10a; - 24 Solution: Compare a; 2 — 9a; — 36 with ax 2 + bx + c and note that a = 1, b = —9, and c = —36. Calculate ac = (1)(— 36), so ac = —36. At this point, some readers might ask "What if I start listing the ordered pairs and I see the pair I need? Do I need to continue listing the remaining pairs? 416 CHAPTER 6. FACTORING Answer: (a; + 12) (x - 2) The answer is "No." In this case, we start listing the integer pairs whose product is ac = —36, but are mindful that we need an integer pair whose sum is b = — 9. The integer pair 3 and —12 has a product equaling ac = —36 and a sum equaling b = — 9. 1,-36 2,-18 -12 Note how we ceased listing ordered pairs the moment we found the pair we needed. Next, replace the middle term —9a; with a sum of like terms using the circled pair. Factor by grouping. x 2 -9x - 36 = x 2 +3x - Ylx - 36 x 2 -9x - 36 = x(x + 3) - 12(ar + 3) = (x- 12)(x + 3) Use the FOIL shortcut to check your answer. (x + 3)(x- 12) F ^2 o Ylx I 3.7' L 36 Combining like terms, (x + 3) (a; — 12) Our solution checks. 9a; — 36, the original trinomial. □ You Try It! Factor: x 2 - Ylx + 35 -12 Speeding Things Up a Bit Readers might again ask "Is it a coincidence that the circled pair seemed to 'drop in place' in the resulting factoriztion (x — 12) (a: + 3) ?" The answer is "No," it is not a coincidence. Provided the leading coefficient of the trinomial ax 2 + bx + c is a = 1, you can always "drop in place" the circled pair in order to arrive at the final factorization, skipping the factoring by grouping. Some readers might also be asking "Do I really have to list any of those ordered pairs if I already recognize the pair I need?" The answer is "No!" If you see the pair you need, drop it in place. EXAMPLE 5. Factor: 5a; - 24. Solution: Compare x 2 — 5a; — 24 with ax 2 + bx + c and note that a = 1, b = — 5, and c = —24. Calculate ac = (1)(— 24), so ac = —24. Now can you think of an integer pair whose product is ac = —24 and whose sum is 6 = —5? For some, the required pair just pops into their head: —8 and 3. The product of these 6.3. FACTORING AX 2 + BX + C WHEN A=l 417 two integers is —24 and their sum is —5. "Drop" this pair in place and you are done. x 2 -5x-24 = {x-8)(x + 3) Use the FOIL shortcut to check your answer. FOIL (x-8)(x + 3) = x 2 + 3x - 8x - 24 Combining like terms, (a: — 8)(x + 3) = x 2 — 5x — 24, the original trinomial. Our solution checks. Answer: (x — 7)(x — 5) The "Drop in Place" technique of Example 5 allows us to revise the ac- method a bit. Revised ac-method. Compare the given polynomial with the standard form ax 2 + bx + c, determine the coefficients a, b, and c, then determine a pair of integers whose product equals ac and whose sum equals b. You then have two options: 1. Write the middle term as a product of like terms using the ordered pair whose product is ac and whose sum is b. Complete the factorization process by factoring by grouping. 2. (Only works if a = 1.) Simply "drop in place" the ordered pair whose product is ac and whose sum is b to complete the factorization process. Note: We'll see in Factoring ax 2 + bx + c when a ^ 1 on page J^21 why this "drop in place" choice does not work when a/1. Readers are strongly encouraged to check their factorization by determining the product using the FOIL method. If this produces the original trinomial, the factorization is correct. Nonlinear Equations Revisited The ability to factor trinomials of the form ax 2 + bx + c, where a = 1, increases the number of nonlinear equations we are now able to solve. □ You Try It! EXAMPLE 6. Solve the equation x 2 = 2x + 3 both algebraically and Solve the equation graphically, then compare your answers. x 2 = —3a; + 4 both algebraically and graphically, then compare your answers. 418 CHAPTER 6. FACTORING Solution: Because there is a power of x larger than one, the equation is nonlinear. Make one side zero. x x 2 - 2x -2x- 3 2x 3 Original equation. Subtract 2x from both sides. Subtract 3 from both sides. Compare x 2 — 2x — 3 with ax 2 + bx + c and note that a = 1, b = —2 and c = —3. We need an integer pair whose product is ac = —3 and whose sum is 6 = — 2. The integer pair 1 and —3 comes to mind. "Drop" these in place to factor. 0+1)0-3) = Factor. We have a product that equals zero. Use the zero product property to complete the solution. or x + 1 =0 x = -l Thus, the solutions of x 2 = 2x + 3 are x x-3 = x = 3 1 and x = 3. Graphical solution. Load each side of the equation x 2 = 2x + 3 into the Y= menu of your graphing calculator, y = x 2 in Yl, y = 2x + 3 in Y2 (see Figure 6.10). Select 6:ZStandard from the ZOOM menu to produce the image at the right in Figure 6.10. Plotl PlotE Plots ^V£B2*X+3 Figure 6.10: Sketch y = x 2 and y = 2x + 3. One of the intersection points is visible on the left, but the second point of intersection is very near the top of the screen at the right (see Figure 6.10). Let's extend the top of the screen a bit. Press the WINDOW button and make adjustments to Ymin and Ymax (see Figure 6.11), then press the GRAPH button to adopt the changes. Note that both points of intersection are now visible in the viewing window (see Figure 6.11). To find the coordinates of the points of intersection, select 5:intersect from the CALC menu. Press the ENTER key to accept the "First curve," press ENTER again to accept the "Second curve," then press ENTER again to accept the current position of the cursor as your guess. The result is shown in the image on the left in Figure 6.12. Repeat the process to find the second point of intersection, only when it comes time to enter your "Guess," use the right-arrow key to move the cursor closer to the second point of intersection than the first. 6.3. FACTORING AX 2 + BX + C WHEN A = 1 419 l.i.lIHDOI.1 Xmin= -10 Xmax= 10 y.scl = 1 Vnin= -5 Vmax= 15 Yscl = 1 l¥.res= I Figure 6.11: Adjusting the view. Intersection K=-l / V=l \v Intersection K=3 / Y=9 Figure 6.12: Use 5:intersect from the CALC menu to find points of intersection. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with x and y, respectively (see Figure 6.13). • Place your WINDOW parameters at the end of each axis (see Figure 6.13). • Label each graph with its equation (see Figure 6.13). • Drop dashed vertical lines through each point of intersection. Shade and label the x-values of the points where the dashed vertical line crosses the x-axis. These are the solutions of the equation x 2 = 2x + 3 (see Figure 6.13). Finally, note how the graphical solutions of a: 2 = 2x + 3, namely x = — 1 and x = 3, match the solutions found using the algebraic method. This is solid evidence that both methods of solution are correct. However, it doesn't hurt to check the final answers in the original equation, substituting —1 for x and 3 for x. (-1) 2 1 22 + 3 2(-l) + 3 -2 + 3 and x (3) 2 9 2x + 3 2(3) + 3 6 + 3 Because the last two statements are true statements, the solutions x x = 3 check in the original equation x 2 = 2x + 3. -1 and Answer: —4, 1 TV Intersection K=-4 Y=lfiX \J Intersection K=l X-IK □ 420 CHAPTER 6. FACTORING Figure 6.13: Reporting your graphical solution on your homework. You Try It! Solve the equation x 2 - 2\x + 90 = both algebraically and graphically, then compare your answers. EXAMPLE 7. Solve the equation x 2 graphically, then compare your answers. 4x — 96 = both algebraically and Solution: Because there is a power of x larger than one, the equation x 2 — Ax — 96 = is nonlinear. We already have one side zero, so we can proceed with the factoring. Begin listing integer pairs whose product is ac = —96, mindful of the fact that we need a pair whose sum is b = —4. 1,-96 2,-48 3,-32 4,-24 6,-16 -12 Note that we stopped the listing process as soon as we encountered a pair whose sum was b = —4. "Drop" this pair in place to factor the trinomial. x z - Ax - 96 (x + 8)(x- 12) Original equation. Factor. 6.3. FACTORING AX 2 + BX + C WHEN A = 1 421 We have a product that equals zero. Use the zero product property to complete the solution. x + : a; -12 = cc = 12 Thus, the solutions of x 2 Ax are x and x = 12. Graphical solution. Load the equation y = x 2 — Ax — 96 in Yl in the Y= menu of your graphing calculator (see Figure 6.14). Select 6:ZStandard from the ZOOM menu to produce the image at the right in Figure 6.14. Plotl PlotE Plots WiBX A 2-4*X-96 -■■Vi = Figure 6.14: Sketch the graph of y Ax - 96. When the degree of a polynomial is two, we're used to seeing some sort of parabola. In Figure 6.14, we saw the graph go down and off the screen, but we did not see it turn and come back up. Let's adjust the WINDOW parameters so that the vertex (turning point) of the parabola and both ^-intercepts are visible in the viewing window. After some experimentation, the settings shown in Figure 6.15 reveal the vertex and the ^-intercepts. Press the GRAPH button to produce the image at the right in Figure 6.15. WINDOW Xnin=-15 Xnax=2@ Xscl=5 Ymin=-150 Vnax=15@ Yscl=50 Figure 6.15: Adjusting the view. Note that both ^-intercepts of the parabola are now visible in the viewing window (see Figure 6.15). To find the coordinates of the x-intercepts, select 2:zero from the CALC menu. Use the left- and right-arrow keys to move the cursor to the left of the first x- intercept, then press ENTER to mark the "Left bound." Next, move the cursor to the right of the first x-intercept, then press ENTER to mark the "Right bound." Press ENTER to accept the current position of the cursor as your "Guess." The result is shown in the image on the left in Figure 6.16. Repeat the process to find the coordinates of the second x-intercept. The result is show in the image on the right in Figure 6.16. 422 CHAPTER 6. FACTORING \ V=0 \l V=0 Figure 6.16: Use 2:zero from the CALC menu to find the ^-intercepts. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with x and y, respectively (see Figure 6.17). • Place your WINDOW parameters at the end of each axis (see Figure 6.17). • Label the graph with its equation (see Figure 6.17). • Drop dashed vertical lines through each x-intercept. Shade and label the x- values of the points where the dashed vertical line crosses the x-axis. These are the solutions of the equation x 2 — Ax — 96 = (see Figure 6.17). 150 -150 y = x 2 - Ax - 96 -»■ x 20 Figure 6.17: Reporting your graphical solution on your homework. Finally note how the graphical solutions of x 2 — Ax — 96 = 0, namely x = —8 and x = 12, match the solutions found using the algebraic method. This is solid evidence that both methods of solution are correct. However, it doesn't 6.3. FACTORING AX 2 + BX + C WHEN A = 1 423 hurt to check the final answers in the original equation, substituting —8 for x Answer: 6, 15 and 12 for x. x z - Ax - 96 = (-8) 2 -4(- 64 + 32 and x 2 - Ax - 96 = (12) 2 -4(12) -96 = 144 - 48 - 96 = Because the last two statements are true statements, the solutions x = —8 and x = 12 check in the original equation x 2 — Ax — 96 — 0. □ 424 CHAPTER 6. FACTORING **> t* t* Exercises ■** ■*$ •** In Exercises 1-6, factor the quadratic polynomial. 1. a; 2 + 7a; -18 4. x 2 + Ylx + 27 2. x 2 + 18a; + 80 5. a; 2 + 14a; + 45 3. a; 2 -10a; + 9 6. a; 2 + 9a; + 20 In Exercises 7-12, factor the quadratic polynomial. 7. a; 2 - 16a; + 39 10. a; 2 - 22a; + 57 8. a; 2 - 16a; + 48 11. a; 2 - 25a; + 84 9. a; 2 - 26a; + 69 12. a; 2 + 13a; - 30 In Exercises 13-18, factor the quadratic polynomial. 13. a; 2 - 13a; + 36 16. a; 2 - 17a; + 66 14. a; 2 + x -12 17. a; 2 - 4a; - 5 15. a; 2 + 10a; + 21 18. a; 2 - 20a; + 99 In Exercises 19-24, use an algebraic technique to solve the given equation. 19. x 2 = -7x + 30 22. a; 2 = x + 72 20. a; 2 = -2a; + 35 23. x 2 = -15x - 50 21. a; 2 = -lice- 10 24. a; 2 = -7x - 6 In Exercises 25-30, use an algebraic technique to solve the given equation. 28. 80 = a; 2 - 16a; 29. 56 = a; 2 + 10a; 30. 66 = a; 2 + 19a; 25. 60 = a; 2 + 11a; 26. -92 = ar - -27a; 27. -11 = X - - 12a; 6.3. FACTORING AX 2 + BX + C WHEN A=l 425 In Exercises 31-36, use an algebraic technique to solve the given equation. 31. x 2 + 20 = -12a; 34. x 2 + 6 = 5a; 32. x 2 - 12 = 11a; 35. x 2 + 8 = -6x 33. x 2 - 36 = 9a; 36. a; 2 + 77 = 18a; In Exercises 37-40, perform each of the following tasks: i) Use a strictly algebraic technique to solve the given equation. ii) Use the 5:intersect utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown in Example 6. 37. x 2 = x + 12 39. x 2 + 12 = 8a: 38. x 2 = 20 - x 40. x 2 + 7 = 8x In Exercises 41-44, perform each of the following tasks: i) Use a strictly algebraic technique to solve the given equation. ii) Use the 2:zero utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown in Example 7. 41. x 2 -6x- 16 = 43. x 2 + 10a; - 24 = 42. x 2 + 7a;- 18 = 44. x 2 - 9x - 36 = it- it- it- Answers •*$ •** •** 15. (a; + 3)(a; + 7) 17. (x + l)(x-5) 19. x = 3, -10 21. x = -1,-10 23. x = -5,-10 25. x = 4, -15 27. x = 1,11 29. a; = 4, -14 1. (x-2)(x + 9) 3. (x- l)(x-9) 5. (a; + 5)(a; + 9) 7. (x-3)(x- 13) 9. (x-3)(x-23) 11. (x-4)(x-21) 13. (x-4)(x-9) 426 CHAPTER 6. FACTORING 31. a; = -2,-10 39. a; = 2, 6 33 ^ = - 3 ' 12 41., = 8, -2 35. x = -2,-4 43. x = -12,2 37. x = -3,4 6.4. FACTORING AX 2 + BX + C WHEN A ^ 1 427 6.4 Factoring ax 2 + bx + c when o/l In this section we continue to factor trinomials of the form ax 2 + bx + c. In the last section, all of our examples had a = 1, and we were able to "Drop in place" our circled integer pair. However, in this section, a/ 1, and we'll soon see that we will not be able to use the "Drop in place" technique. However, readers will be pleased to learn that the ac-method will still apply. EXAMPLE 1. Factor: 2a; 2 - 7x - Solution: We proceed as follows: 15. 1. Compare 2a; 2 — 7a; — 15 with aa; 2 + bx + c and identify a = 2, b = —7, and c = —15. Note that the leading coefficient is a = 2, so this case is different from all of the cases discussed in Section 6.3. 2. Calculate ac. Note that ac = (2)(— 15), so ac = —30. 3. List all integer pairs whose product is ac = —30. 1. -30 -1,30 2, -15 -2,15 3, -10 -3,10 ■5. -6 -5,6 You Try It! Factor: 3a; 2 + 13a; + 14 4. Circle the ordered pair whose sum is -7. 1. 2, -30 -15 -1,30 -2,15 :'.. -10 -3,10 5, -6 -5,6 Note that if we "drop in place" our circled ordered pair, (x + 3)(x— 10) 7^ 2a; 2 — 7x — 15. Right off the bat, the product of the terms in the "First" position does not equal 2a; 2 . Instead, we break up the middle term of 2a; 2 — 7a; — 15 into a sum of like terms using our circled pair of integers 3 and -10. 2x 2 -7a; - 15 = 2a; 2 +3a; - 10a; - 15 Now we factor by grouping. Factor x out of the first two terms and —5 out of the second two terms. x (2a; + 3) - 5 (2a; + 3) Now we can factor out (2a; + 3). (a;-5)(2a; + 3) 428 CHAPTER 6. FACTORING Answer: (x + 2)(3x + 7) 6. Use the FOIL shortcut to mentally check your answer. To multiply (x — 5) (2a; + 3), use these steps: • Multiply the terms in the "First" positions: 2x 2 . • Multiply the terms in the "Outer" and "Inner" positions and com- bine the results mentally: 3x — Wx = —7x. • Multiply the terms in the "Last" positions: —15. That is: (x-5)(2sc + 3) F 2x 2 O 3.r 10a; L 15 Combining like terms, (x — 5)(2x + 3) = 2a; 2 — 7x — 15, which is the original trinomial, so our solution checks. Note that if you combine the "Outer" and "Inner" products mentally, the check goes even faster. □ You Try It! Factor: 2a: 2 - 9a; + 10 Speeding Things Up a Bit Some readers might already be asking "Do I really have to list all of those ordered pairs if I already see the pair I need?" The answer is "No!" If you see the pair you need, use it to break up the middle term of the trinomial as a sum of like terms. EXAMPLE 2. Factor: 3a; 2 - 7x - 6. Solution: Compare 3a; 2 — 7a; — 6 with ax 2 + bx + c and note that a = 3, b = —7, and c = —6. Calculate ac = (3) (—6), so ac = —18. Now can you think of an integer pair whose product is ac = —18 and whose sum is b = —7? For some, the pair just pops into their head: 2 and —9. Break up the middle term into a sum of like terms using the pair 2 and —9. 3x 2 -7a;-6 = 3a; 2 +2a; - 9a; - 6 = x (3a; + 2) - 3 (3a; + 2) = (x-3)(3x + 2) Use the FOIL shortcut to check your answer F (x — 3)(3a; - 3x 2 — 7a; = 2x -9x. Factor by grouping Factor out (3a; + 2) / L 2a; - 9x - 6 6.4. FACTORING AX 2 + BX + C WHEN A ^ 1 429 Combining like terms, (x — 3)(3a; + 2) = 3a; 2 — 7x — 6, the original trinomial. Our solution checks. Note that if you combine the "Outer" and "Inner" products mentally, the check goes even faster. Answer: (x — 2)(2x — 5) □ On the other hand, some readers might be saying "Well, the needed ordered pair is not popping into my head. Do I have a way of cutting down the work?" The answer is "Yes!" As you are listing the ordered pairs whose product equals ac, be mindful that you need the ordered pair whose sum is b. If you stumble across the needed pair, stop the listing process and "drop" your ordered pair in place. EXAMPLE 3. Factor: 3x 2 - 33x + 54. Solution: Compare 3a; 2 — 33x + 54 with ax 2 + bx + c and note that a = 3, b = —33, and c = 54. Calculate ac = (3)(54), so ac = 162. Ouch! That's a big number! However, start listing the integer pairs whose product is ac = 162, but be mindful that you need an integer pair whose sum is b = —33. You Try It! Factor: 5x 2 - 35x - 40 1,162 2,81 3,54 6,27 -6,-27 As soon as we wrote down the pair 6 and 27, our mind said "the sum of 6 and 7 is 33." However, we need the sum to equal b = —33, so we boxed —6 and — 27 instead. Next, we break up the middle term into a sum of like terms using our circled pair. 3x -33x - 54 = 3x 2 -6x- 27a; - 54 = 3x(x-2)-27(x-2) = (3x-27)(x-2) — 33a; = — 6x — 27x. Factor by grouping. Factor out (x — 2). Oh-oh! Now we realize we can factor 3 out of each term in the first factor! = 3(x-9)(x-2) We missed taking out the GCF! Let's try again, only this time let's do what we are always supposed to do in the first step: Factor out the GCF. 3a^ 33x + 54 = 3(x 2 - llx + 18) 430 CHAPTER 6. FACTORING Answer: 5 (a; — 8)(x + 1) Comparing x 2 — l\x + 18 with ax 2 + bx + c, we see that a = 1, b = — 11, and c = 18. We need an integer pair whose product is ac = 18 and whose sum is b = — 11. Note that these numbers are considerably smaller than the numbers we had to deal with when we forgot to first factor out the GCF. Because the numbers are smaller, the integer pair —9 and —2 easily conies to mind. Furthermore, because a = 1, we can factor x 2 — 11a; + 18 by simply dropping the integer pair —9 and —2 in place. 3(x 2 - Ux + 18) = 3(x - 9)(x - 2) A far simpler solution! □ In Example 3, we saw how much more difficult we made the problem by forgetting to first factor out the greatest common factor (GCF). Let's try not to make that mistake again. First rule of factoring. The first step in factoring any polynomial is to factor out the greatest common factor. You Try It! Factor: 12a; 4 + 2a; 3 - 30a; 2 EXAMPLE 4. Factor: 30a; 3 - 21a; 2 - 18a;. Solution: Note that the GCF of 30a; 3 , 21a; 2 , and 18a; is 3a;. Factor out this GCF. — 3a; ■ 7a; - 7x - 6) c and note that a 30x^ - 2\x z - 18a; = 3a; ■ lOa;^ - 3a- ■ Ix - 3a; • 6 = 3a;(10a; 2 Next, compare 10a: 2 — 7a; — 6 with aa: 2 + bx - and c = — 6. Start listing the integer pairs whose product is ac mindful that you need an integer pair whose sum is b = — 7. 1,-60 2,-30 3,-20 4,-15 10, b = -7, —60, but be 12 Break up the middle term into a sum of like terms using our circled pair. Hence, 30ar 3a;(10a; 2 - 7x - 6) = 3a;(10a; 2 + 5a; - 12a;- 18) = 3a;[5a;(2a;+ 1) - 6(2a; + 1)] = 3a;(5a;-6)(2a; + l) 21a; 2 - 18a; = 3a;(5a;-6)(2x — 7a; = 5a; — 12a;. Factor by grouping. Factor out a 2a; + 1 . 6.4. FACTORING AX 2 + BX + C WHEN A ^ 1 431 Check: First, use the FOIL shortcut to multiply the two binomial factors, then distribute the monomial factor. 3x(5x - 6) (2x + 1) = 3x(10a; 2 - 7x - 6) Apply the FOIL shortcut. = 30a; 3 - 2 la; 2 - 18a; Distribute the 3a;. Because this is the original polynomial, the solution checks. Answer: 2x 2 (3x + 5)(2x - 3) □ Nonlinear Equations Revisited Let's use the factoring technique of this chapter to solve some nonlinear equa- tions. You Try It! EXAMPLE 5. Solve the equation 2a; 2 graphically, then compare your answers. 13a; — 20 both algebraically and Solve the equation 5a; 2 = 12a; + 9 both algebraically and graphically, Solution: Because there is a power of x larger than one, the equation is nonlinear. Make one side equal to zero. 2a; = 13x — 20 Original equation. 2a; — 13a; + 20 = Make one side zero. Compare 2a; 2 — 13a; + 20 with ax 2 + bx + c and note that a = 2, b = — 13 and c = 20. We need an integer pair whose product is ac = 40 and whose sum is b = —13. The integer pair —5 and —8 comes to mind. Write the middle term as a sum of like terms using this pair. then compare your answers. 2a; 2 - 5a; - 8a; + 20 = x(2x - 5) - 4(2x - 5) = (x-4)(2a;-5) = — 13x = — 5x — 8a;. Factor by grouping. Factor out 2x — 5. We have a product that equals zero. Use the zero product property to complete the solution. Thus, the solutions of 2a; a;-4 = or 2x — 5 = x = 4 2a; = 5 5 X= 2 2a; 2 = 13a; - - 20 are x = 4 and x 5/2. Graphical solution. Load each side of the equation 2x 2 = 13a; — 20 into the Y= menu of your graphing calculator, y = 2x 2 in Yl, y = 13a; — 20 in Y2 432 CHAPTER 6. FACTORING (see Figure 6.18). Select 6:ZStandard from the ZOOM menu to produce the image on the left in Figure 6.18. However, even after adjusting the WINDOW parameters (Xmin = —10, Xmax = 10, Ymin = —10, and Ymax = 60), the image resulting from pushing the GRAPH button (see the image on the right in Figure 6.18) does not clearly show the two points of intersection. Figure 6.18: Sketch y = 2x 2 and y = 13x - 20. Let's switch our strategy and work with the equation 2x 2 - 13x + 20 = instead. Load y = 2x 2 — 13x + 20 into Yl in the Y= menu, then select 6:ZStandard from the ZOOM menu to produce the image at the right in Figure 6.19. Plotl PlotE Plots WiB2*X A 2-13*X+2 Wh = Figure 6.19: Sketch y = 2x 2 - 13a; + 20. To find the solutions of 2a; 2 — 13a; + 20 = 0, we must identify the a;-intercepts of the graph in Figure 6.18. Select 2:zero from the CALC menu, then move the left- and right=arrows to move the cursor to the left of the first x-intercept. Press ENTER to mark the "Left bound," then move the cursor to the right of the a;-intercept and press ENTER to mark the "Right bound." Finally, press ENTER to use the current position of the cursor for your "Guess." The result is shown in the image on the left in Figure 6.20. Repeat the process to find the rightmost x-intercept. The result is shown in the image on the right in Figure 6.20. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with x and y, respectively (see Figure 6.21). 6.4. FACTORING AX 2 + BX + C WHEN A ^ 1 433 IV Z*K* K=2.£ V=0 M Z*K* Figure 6.20: Use 2:zero from the CALC menu to find the x- intercepts. • Place your WINDOW parameters at the end of each axis (see Figure 6.21). • Label the graph with its equation (see Figure 6.21). • Drop dashed vertical lines through each ^-intercept. Shade and label the a;-values of the points where the dashed vertical line crosses the x- axis. These are the solutions of the equation 2a: 2 — 13a; + 20 = (see Figure 6.21). 10 -10 y = 2x 2 - 13a; + 20 -*- x 10 -10 Figure 6.21: Reporting your graphical solution on your homework. Finally, note how the graphical solutions of 2a; 2 — 13a: + 20 = 0, namely x = 2.5 and x = 4, match the solutions x = 5/2 and x = 4 found using the algebraic method. This is solid evidence that both methods of solution are correct. However, it doesn't hurt to check the final answers in the original equation, 434 CHAPTER 6. FACTORING Answer: —3/5, 3 V \ V=0 ...A \ K=3 V=0 substituting 5/2 for x and 4 for x. and 2aT = 13a; - 20 2(4) 2 = 13(4) - 20 2(16) = 13(4) -20 32 = 52 - 20 Because the last two statements are true statements, the solutions x = 5/2 and x = 4 check in the original equation 2a; 2 = 13a; — 20. □ 2x z = 13a; - 20 2 <$) = 13 (I) " 2 ° 2 :d = 13(0-20 25 65 40 2 2 2 You Try It! Solve the equation 4a; d 14a; both algebraically and graphically, then compare your answers. EXAMPLE 6. Solve the equation 2x 3 graphically, then compare your answers. 28a; both algebraically and Solution: Because there is a power of x larger than one, the equation is nonlinear. Make one side equal to zero. 2a; + x = 28a; Original equation. 2x + x — 28a; = Make one side zero. Note that the GCF of 2a; 3 , a; 2 , and 28a; is x. Factor out x. x(2x 2 + x - 28) = Factor out the GCF. Compare 2a; 2 + a;— 28 with aa; 2 + 6a: + c and note that a = 2, b = 1 and c = —28. We need an integer pair whose product is ac = —56 and whose sum is b = 1. The integer pair —7 and 8 comes to mind. Write the middle term as a sum of like terms using this pair. a;(2a; 2 - 7x + 8x - 28) = a;[a;(2a;-7) + 4(2a;-7)] =0 a;(a; + 4)(2a;-7) = x = — 7x + 8x. Factor by grouping. Factor out 2a; — 7. We have a product of three factors that equals zero. By the zero product property, at least one of the factors must equal zero. a; + 4 x 2a; - 7 = 2a; = 7 7 2 x 6.4. FACTORING AX 2 + BX + C WHEN A ^ 1 435 Thus, the solutions of 2a; 3 28a; are x = 0, x = —4, and x = 7/2. Graphical solution. Rather than working with 2a; 3 28x, graphing each side separately and finding where the graphs intersect, we will work instead 28a; = 0, locating where the graph of y = 2a; 3 + x 2 — 28a; ; 2 — 28a; into Yl in the Y= menu, then with 2ar crosses the a:-axis. Load y = 2x + : select 6:ZStandard from the ZOOM menu to produce the image at the right in Figure 6.22. Plotl PlotE Plots WiB2*X A 3+>T-2-28 Wh = If i I Figure 6.22: Sketch y = 2x 3 28a;. In the image at the right in Figure 6.22, we saw the graph rise from the bottom of the screen, leave the top of the screen, return and leave via the bottom of the screen, and then finally return and leave via the top of the screen. Clearly, there are at least two turning points to the graph that are not visible in the current viewing window. Set the WINDOW settings as shown in the image on the left in Figure 6.23, then push the GRAPH button to produce the image on the right in Figure 6.23. Note that this window now shows the a;-intercepts as well as the turning points of the graph of the polynomial. WINDOW Xnin=-10 Xnax=10 Kscl=l Ymin=-100 Vnax=100 Yscl=10 Figure 6.23: Adjusting the view so that the turning points of the polynomial are visible. To find the solutions of 2a; 28a; = 0, we must identify the x-intercepts of the graph in Figure 6.23. Select 2:zero from the CALC menu, then use the left- and right-arrow keys to move the cursor to the left of the first a;- intercept. Press ENTER to mark the "Left bound," then move the cursor to the right of the a;-intercept and press ENTER to mark the "Right bound." Finally, press ENTER to use the current position of the cursor for your "Guess." The result is shown in the first image on the left in Figure 6.24. Repeat the process to find the remaining a;-intercepts. The results are shown in the next two images in Figure 6.24. 436 CHAPTER 6. FACTORING / f' " Z*K* J k=-h r KJ Z*K* J k=o r v=o J r ' ' ■ Z*K* J K=3.E r jj6 Figure 6.24: Use 2:zero from the CALC menu to find the x- intercepts. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with x and y, respectively (see Figure 6.25). • Place your WINDOW parameters at the end of each axis (see Figure 6.25) . • Label the graph with its equation (see Figure 6.25). • Drop dashed vertical lines through each x-intercept. Shade and label the x-values of the points where the dashed vertical line crosses the pr- axis. These are the solutions of the equation 2a; 3 + x 2 — 28a; = (see Figure 6.25). y = 2x 3 28a; Answer: -2, 0, 7/4 Z*K* K=-£ Z*K0 K=1.7E Z*K0 J K=0 I V=0 100' 10 -4/ -100' V /f 3.5 1 10 Figure 6.25: Reporting your graphical solution on your homework. x = 0, and x using the algebraic method 3_L/r.2 OV ,- II ,,-,,,.,,1,- ,- /J. 3.5, match the solutions x = —4, x = 0, and x = 7/2 found Finally note how the graphical solutions of 2x A -\-x 2 — 28a; = 0, namely x □ 6.4. FACTORING AX 2 + BX + C WHEN A^l 437 **.**.**. Exercises -*s •*& >•$ In Exercises 1-6, factor the quadratic polynomial. 1. 6x 2 + 13x - 5 4. 6a; 2 - 23a; + 7 2. 3a; 2 - 19a; + 20 5. 3a; 2 + 19a; + 28 3. 4a; 2 -x -3 6. 2a; 2 - 9a; - 18 In Exercises 7-12, factor the quadratic polynomial. 7. 12a; 2 - 23a; + 5 10. 4a; 2 + 19a; + 21 8. 8a; 2 + 22a; + 9 11. 3a; 2 + 4a; - 32 9. 6a; 2 + 17a; + 7 12. 4a; 2 + x - 14 In Exercises 13-18, factor the quadratic polynomial. 13. 3a; 2 + 19a; + 28 16. 4a; 2 - x - 14 14. 6a; 2 + a;-l 17. 6a; 2 - 11a; - 7 15. 4a; 2 -21a; + 5 18. 2a; 2 - 17a; + 21 In Exercises 19-26, factor the trinomial. 19. 16a; 5 - 36a; 4 + 14a; 3 23. 6a; 4 - 33a; 3 + 42a; 2 20. 12a; 4 - 20a; 3 + 8a; 2 24. 15a; 3 - 10a; 2 - 105a; 21. 36a; 4 - 75a; 3 + 21a; 2 25. 16a; 4 - 36a; 3 - 36a; 2 22. 6a; 4 - 10a; 3 - 24a; 2 26. 40a; 4 - 10a; 3 - 5a; 2 In Exercises 27-38, use an algebraic technique to solve the given equation. 27. 4a; 2 = -x + 18 30. 2a; 2 - 20 = -3a; 28. 2a; 2 = 7x-3 31. 3a; 2 + 30 = 23a; 29. 3a; 2 + 16 = -14a; 32. 6a; 2 - 7 = -11a; 438 CHAPTER 6. FACTORING 33. -7x - 3 = -6a; 2 36. -23a; + 7 = -6a; 2 34. 13a; -45 = -2a; 2 37. 6x 2 = -25a; + 9 35. 26a;- 9 = -3a; 2 38. 2.x 2 = 13a; + 45 In Exercises 39-42, perform each of the following tasks: i) Use a strictly algebraic technique to solve the given equation. ii) Use the 2:zero utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown in Example 5. 39. 2a; 2 - 9a; - 5 = 41. 4a; 2 - 17a; - 15 = 40. 2a; 2 + x - 28 = 42. 3a; 2 + 14a; - 24 = In Exercises 43-46, perform each of the following tasks: i) Use a strictly algebraic technique to solve the given equation. ii) Use the 2:zero utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown in Example 6. 43. 2a; 3 = 3a; 2 + 20a; 45. 10a; 3 + 34a; 2 = 24a; 44. 2a; 3 = 3x 2 + 35a; 46. 6x 3 + 3a; 2 = 63a; $*-$*>£»■ Answers •*$ -*$ •*$ 1. (2a; + 5)(3a;- 1) 3. [x- l)(4a; + 3) 5. ^a; + 4)(3a; + 7) 7. ^3a;-5)(4a;- 1) 9. ^2a; + l)(3a; + 7) 11. ^a; + 4)(3a;-8) 13. (a; + 4)(3a; + 7) 15. [x-5)(ix- 1) 17. (3a; -7) (2a; + 1) 19. 2a; 3 (2a;- l)(4x- 7) 21. 3a; 2 (3a;- l)(4a;- 7) 23. 3a; 2 (a;-2)(2a;-7) 25. 4a; 2 (x- 3)(4a; + 3) 27. 9 x = 2,-- 29. x = -2, — ' 3 31. , 5 x = o, - ' 3 1 33. x = --. 3 3 2 35. x = -9, 1 3 37.,= !,- 9 ~2 6.4. FACTORING AX 2 + BX + C WHEN A^\ 439 39. x = -1/2,5 41. x = -3/4,5 43. x = 0,-5/2, 4 45. a; = 0,7/2, -5 440 CHAPTER 6. FACTORING 6.5 Factoring Special Forms In this section we revisit two special product forms that we learned in Chapter 5, the first of which was squaring a binomial. Perfect Square Trinomials To square a binomial such as (a + b) 2 , proceed as follows: 1. Square the first term: a 2 2. Multiply the first and second term, then double: 2ab 3. Square the last term: b 2 You Try It! Expand: (5a + 2b) 2 Answer: 25a 2 + 20ab + 46 2 EXAMPLE 1. Expand: (2x + 3y) 2 Solution: Using the pattern (a + b) 2 = a 2 + 2ab + b 2 , we can expand (2x + 3y) 2 as follows: (2x + 3j/) 2 = (2x) 2 + 2(2x)(3y) + (3y) 2 = Ax + Qxy + 9y Note how we square the first and second terms, then produce the middle term of our answer by multiplying the first and second terms and doubling. □ You Try It! Expand: (2s 3 - lif EXAMPLE 2. Expand: (3m 2 - 5v 2 ) 2 Solution: Using the pattern (a—b) 2 = a 2 — 2ab+b 2 , we can expand (3w 2 — 5i> 2 ) 2 as follows: (3u 2 - 5v 2 ) 2 = (3w 2 ) 2 - 2(3u 2 )(5v 2 ) + (5m 2 ) 2 = 9m 4 - 30m 2 m 2 + 25m 4 6.5. FACTORING SPECIAL FORMS 441 Note that the sign of the middle term is negative this time. The first and last terms are still positive because we are squaring. Answer: At 28sH + 49? Once you've squared a few binomials, it's time to do all of the work in your head, (i) Square the first term; (ii) multiply the first and second term and double the result; and (hi) square the second term. You Try It! EXAMPLE 3. Expand each of the following: a) (2y - 3) 2 b) (4a - 36) 2 c) {x 3 + 5) 2 Solution: Using the pattern (a ± o) 2 = a 2 ± lab + 6 2 , we expand each binomially mentally, writing down the answer without any intermediate steps. a) (2j/ - 3) 2 = 4y 2 - Yly + 9 b) (4a - 3b) 2 = 16a 2 - 24a& + 9b 2 c) (x 3 + 5) 2 = x 6 + 10a; 3 + 25 Expand: (5z 4 - 3) : Answer: 25a You Try It! EXAMPLE 4. Factor each of the following trinomials: a) 4y 2 - 12y + 9 b) 16a 2 - 24a6 + 9b 2 c) x 6 + 10a; 3 + 25 Solution: Because of the work already done in Example 3, it is a simple task to factor each of these trinomials. a) Ay 2 - Yly + 9 = (2y - 3) 2 b) 16a 2 - 24a6 + 96 2 = (4a - 36) 2 c) x 6 + 10a; 3 + 25 = (a; 3 + 5) 2 Factor: 25a; 8 - 30.x 4 Answer: (5x 4 — 3)" 30a; 4 + 9 □ Now, because factoring is "unmultiplying," it should be a simple matter to reverse the process of Example 3. □ □ 442 CHAPTER 6. FACTORING Each of the trinomials in Example 4 is an example of a perfect square trinomial. Perfect square trinomial. If a trinomial a 2 + 2ab + b 2 is the square of a binomial, as in (a + b) 2 , then the trinomial is called a perfect square trinomial. You Try It! Factor: 16a; 2 + 72a; + 81 List of Squares n n 2 1 1 2 4 3 9 4 16 5 25 6 36 7 49 8 64 9 81 10 100 11 121 12 144 13 169 14 196 15 225 16 256 17 289 18 324 19 361 20 400 21 441 22 484 23 529 24 576 25 625 So, how does one recognize a perfect square trinomial? If the first and last terms of a trinomial are perfect squares, then you should suspect that you may be dealing with a perfect square trinomial. However, you also have to have the correct middle term in order to have a perfect square trinomial. EXAMPLE 5. Factor each of the following trinomials: a) 9a; 2 - 42a; + 49 b) 49a 2 + 70ab + 256 2 c) 4a; 2 - 37a; + 9 Solution: Note that the first and last terms of each trinomial are perfect squares. a) In the trinomial 9a; 2 — 42a; + 49, note that (3a;) 2 = 9a; 2 and 7 2 = 49. Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 9a; 2 — 42a; + 49 factors as follows: 9a; 2 42a; + 49 = (3.x - 7) 2 However, we must check to see if the middle term is correct. Multiply 3a; and 7, then double: 2 (3a;) (7) = 42a; . Thus, the middle term is correct and therefore 9a; 2 42a; + 49 = (3a; - 7)' b) In the trinomial 49a 2 +70a6+256 2 , note that (7a) 2 = 49a 2 and {5b) 2 = 25b 2 . Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 49a 2 + 70a6 + 256 2 factors as follows: 49a 2 + 70a6 + 25o 2 = (7a + 5o) 2 However, we must check to see if the middle term is correct. Multiply 7a and 56, then double: 2(7a)(56) = 70a6. Thus, the middle term is correct and therefore 49a 2 + 70a6 + 256 2 = (7a + 56) 2 . c) In the trinomial 4a; 2 - 37a; + 9, note that (2a;) 2 = 4a; 2 and (3) 2 = 9. Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 4a; 2 — 37a; + 9 factors as follows: 4a; 2 - 37a; + 9 = (2a; - 3) 2 6.5. FACTORING SPECIAL FORMS 443 However, we must check to see if the middle term is correct. Multiply 2x and 3, then double: 2(2a;)(3) = 12.t. However, this is not the middle term of 4a; 2 — 37a; + 9, so this factorization is incorrect! We must find another way to factor this trinomial. Comparing 4a; 2 — 37a; + 9 with ax 2 + bx + c, we need a pair of integers whose product is ac = 36 and whose sum is b = —37. The integer pair —1 and —36 comes to mind. Replace the middle term as a sum of like terms using this ordered pair. 4a; 2 - 37a; + 9 4a; 2 - x - 36a; + 9 a;(4a; - 1) - 9(4a; - (a;-9)(4a;- 1) —37a; = —x — 36a;. 1) Factor by grouping. Factor out 4a; — 1. This example clearly demonstrates how important it is to check the middle term. Answer: (4a; + 9)^ Remember the first rule of factoring! The first rule of factoring. The first step to perform in any factoring problem is factor out the GCF. □ You Try It! EXAMPLE 6. Factor each of the following trinomials: Factor: -4a; d 24a; 2 - 36.x a) 2a; 3 y + 12x 2 y 2 + 18xy 3 b) -4a; 5 + 32a; 4 64x 3 Solution: Remember, first factor out the GCF. In the trinomial 2x 3 y + 12x 2 y 2 + I8xy 3 , we note that the GCF of 2x 3 y, \2x 2 y 2 , and 18xy 3 is 2a;y. We first factor out 2xy. 2x 3 y + 12x 2 y 2 + 18xy 3 = 2xy{x 2 + Qxy + 9y 2 ) We now note that the first and last terms of the resulting trinomial factor are perfect squares, so we take their square roots and factors as follows. = 2a;y(a; + 3y) 2 Of course, the last factorization is correct only if the middle term is correct. Because 2(x)(3y) = Qxy matches the middle term of a; 2 + 6xy + 9y 2 , we do have a perfect square trinomial and our result is correct. Answer: — Ax(x + 3)' 444 CHAPTER 6. FACTORING b) In the trinomial -4a; 5 + 32a: 4 - 64a; 3 , we note that the GCF of 4a; 5 , 32.x 4 , and 64a: 3 is 4a: 3 . We first factor out 4a; 3 . -4a; 5 + 32a; 4 - 64a; 3 = 4a; 3 (-x 2 + 8x - 16) However, the first and third terms of —x 2 + 8a; — 16 are negative, and thus are not perfect squares. Let's begin again, this time factoring out — 4a; . -4a; 5 + 32a; 4 - 64a; 3 = -4a; 3 (x 2 - 8x + 16) This time the first and third terms of a; 2 — 8a; + 16 are perfect squares. We take their square roots and write: = -4a; 3 (a;-4) 2 Again, this last factorization is correct only if the middle term is correct. Because 2 (a;) (4) = 8a;, we do have a perfect square trinomial and our result is correct. □ The Difference of Squares The second special product form we learned in Chapter 5 was the difference of squares. The difference of squares. Here is the difference of squares rule. (a + b)(a-b) = a 2 - b 2 If you are multiplying two binomials which have the exact same terms in the "First" positions and the exact same terms in the "Last" positions, but one set is separated by a plus sign while the other set is separated by a minus sign, then multiply as follows: 1. Square the first term: a 2 2. Square the second term: b 2 3. Place a minus sign between the two squares. 6.5. FACTORING SPECIAL FORMS 445 You Try It! EXAMPLE 7. Expand each of the following: Expand: (4c - 3y)(4a; + 3y) a) (3a; + 5)(3cc - 5) b) (a 3 - 26 3 )(a 3 + 2b 3 ) Solution: We apply the difference of squares pattern to expand each of the given problems. a) In (3x + 5) (3a; — 5), we have the exact same terms in the "First" and "Last" positions, with the first set separated by a plus sign and the second set separated by a minus sign. a) Square the first term: (3a;) 2 = 9x 2 b) Square the second term: 5 2 = 25 c) Place a minus sign between the two squares. Hence: (3a; + 5)(3x-5) = 9a; 2 - 25 b) In (a 3 — 26 3 )(a 3 + 2b 3 ), we have the exact same terms in the "First" and "Last" positions, with the first set separated by a minus sign and the second set separated by a plus sign. a) Square the first term: (a 3 ) 2 = a 6 b) Square the second term: (26 3 ) 2 = 4& 6 c) Place a minus sign between the two squares. Hence: (a 3 -26 3 )(a 3 + 26 3 ) = a 6 -46 6 Answer: 16a; 2 — 9j/ 2 □ Because factoring is "unmultiplying," is should be a simple matter to reverse the process of Example 7. You Try It! EXAMPLE 8. Factor each of the following: Factor: 81a; 2 - 49 a) 9a; 2 - 25 b) a 6 - 46 6 Solution: Because of the work already done in Example 7, it is a simple matter to factor (or "unmultiply" ) each of these problems. a) 9a; 2 -25= (3a; + 5) (3a; - 5) b) a 6 -4b 6 = {a 3 -2b 3 ) (a 3 + 2b 3 ) 446 CHAPTER 6. FACTORING Answer: (9x + 7)(9x - 7) In each case, note how we took the square roots of each term, then separated one set with a plus sign and the other with a minus sign. Because of the commutative property of multiplication, it does not matter which one you make plus and which one you make minus. □ Always remember the first rule of factoring. The first rule of factoring. The first step to perform in any factoring problem is factor out the GCF. You Try It! Factor: 4x — 16.x EXAMPLE 9. Factor: x 3 - 9x Solution: In x 3 — 9x, the GCF of a; 3 and 9x is x. Factor out x. „3 n~. ™/~,2 X 9x = x(x z - 9) Answer: Ax 2 (x + 2) (x - 2) Note that x 2 — 9 is now the difference of two perfect squares. Take the square roots of x 2 and 9, which are x and 3, then separate one set with a plus sign and the other set with a minus sign. = x(x + 3)(x — 3) □ You Try It! Factor: x — 81 Factoring Completely Sometimes after one pass at factoring, factors remain that can be factored further. You must continue to factor in this case. EXAMPLE 10. Factor: x 4 16 Solution: In 16, we have the difference of two squares: (x 2 ) z and 4 2 = 16. First, we take the square roots, then separate one set with a plus sign and the other set with a minus sign. x 4 - 16= (x 2 +4)(x 2 -4) 6.5. FACTORING SPECIAL FORMS 447 Note that x 2 +4 is the sum of two squares and does not factor further. However, x 2 — 4 is the difference of two squares. Take the square roots, x and 2, then separate one set with a plus sign and the other set with a minus sign. (a; 2 +4)(a; + 2)(a--2) Done. We cannot factor further. Answer: (cc 2 +9)(a; + 3)(a;-3) □ Nonlinear Equations Revisited Remember, if an equation is nonlinear, the first step is to make one side equal to zero by moving all terms to one side of the equation. Once you've completed this important first step, factor and apply the zero product property to find the solutions. EXAMPLE 11. Solve for 25a; 2 169 Solution: Make one side equal to zero, factor, then apply the zero product property. 2hx z 25x = 169 Original equation. - 169 = Subtract 169 from both sides. You Try It! Solve for 16x 2 121 Note that we have two perfect squares separated by a minus sign. This is the difference of squares pattern. Take the square roots, making one term plus and one term minus. (5x + 13)(5x- 13) = Use difference of squares to factor. Use the zero product property to complete the solution, setting each factor equal to zero and solving the resulting equations. 5z + 13 = 5x- 13 13 13 ~5~ Hence, the solutions of 25a; 2 = 169 are x = —13/5 and x = 13/5. We encourage readers to check each of these solutions. Answer: -11/4, 11/4 □ 448 CHAPTER 6. FACTORING You Try It! Solve for x: 25a; 2 = 80a; 64 EXAMPLE 12. Solve for x: 49a; 2 + 81 = 126a; Solution: Make one side equal to zero, factor, then apply the zero product property. One can also argue that the only number whose square is zero is the number zero. Hence, one can go directly from (7a -9) 5 to 7x - 9 = 0. Hence, the only solution of 49a; 2 + 81 = 126a; is x = 9/7. 49x 2 n <±9x z - 126a; + 81 126a; Original equation. Subtract 126a; from both sides. Note that the first and last terms of the trinomial are perfect squares. Hence, it make sense to try and factor as a perfect square trinomial, taking the square roots of the first and last terms. (7a;-9) 2 = Factor as a perfect square trinomial. Of course, be sure to check the middle term. Because — 2(7a;)(9) = —126a;, the middle term is correct. Because (7a; — 9) 2 = (7x — 9) (7a; — 9), we can use the zero product property to set each factor equal to zero and solve the resulting equations. Answer: 8/5 7a; - 9 = 9 X= 7 7x - 9 = 9 X= 7 Hence, the only solution of 49a; 2 + 81 = 126a; is x = 9/7. We encourage readers to check this solution. □ You Try It! Solve for 5a; 3 + 36 180a; EXAMPLE 13. Solve for 2a; 3 + 3a; 2 = 50a; + 75 Solution: Make one side equal to zero, factor, then apply the zero product property. 2a; + 3a; = 50a; + 75 Original equation. Make one side zero. 2x 3 + 3a; 2 - 50a; - 75 = This is a four-term expression, so we try factoring by grouping. Factor a; 2 out of the first two terms, and —25 out of the second two terms. a; 2 (2a; + 3)-25(2a; + 3) = (a; 2 -25)(2x + 3) = Factor by grouping Factor out 2a; + 3. 6.5. FACTORING SPECIAL FORMS 449 Complete the factorization by using the difference of squares to factor x 2 — 25. (x + 5)(x — 5) (2x + 3) = Use difference of squares to factor. Finally, use the zero product property. Set each factor equal to zero and solve for x. x + 5 = x = —5 or x-5 = x = 5 or 2x Hence, the solutions of 2x 3 + 3x 2 = 50a; + 75 are x = —5, x = 5, and x We encourage readers to check each of these solutions. -3/2. Answer: —6, 6, 1/5 Let's solve another nonlinear equation, matching the algebraic and graphi- cal solutions. □ You Try It! EXAMPLE 14. Solve the equation x 3 = 4a;, both algebraically and graphi- Solve the equation a; 3 = 16a; cally, then compare your answers. Solution: Note that we have a power of x larger than one, so the equation a; 3 = 4a; is nonlinear. Make one side zero and factor. 4a; a; 3 - 4a; = x(x 2 - 4) = x{x + 2){x-2) = Original equation. Nonlinear. Make one side zero. Factor out GCF. Apply difference of squares. both algebraically and graphically, then compare your answers. a; + 2 = or x - 2 = x = -2 x = 2 4x are x = 0, x = —2, and x = 2. Note that we now have a product of three factors that equals zero. The zero product property says that at least one of these factors must equal zero. Hence, the solutions of a; 3 Graphical solution. Load y = x 3 and y = 4x into Yl and Y2 in the Y= menu of your calculator. Select 6:ZStandard from the ZOOM menu to produce the graph in Figure 6.26. Although the image in Figure 6.26 shows all three points of intersection, adjusting the WINDOW parameters as shown in Figure 6.27, then pressing the GRAPH button will produce a nicer view of the points of intersection, as shown in the figure on the right in Figures 6.27. 450 CHAPTER 6. FACTORING Plotl PlotE PlOtS WiBX A 3 \Y*B4*IK Wj=I ^Vh= We = N.Vfi = \V? = Figure 6.26: Sketching y WINDOW Xmin= _ 5 Xmax=5 Xscl=l Vnin=-15 Vmax=15 Yscl=5 4-Xres=i = a; 3 and y = Ax. Figure 6.27: Adjusting the viewing window. Use the 5:intersect tool from the CALC menu to find the three points of intersection. Press the ENTER key in response to "First curve," then press ENTER again in response to "Second curve," then use the left-arrow key to move your cursor close to the leftmost point of intersection and press ENTER in response to "Guess." The result is shown in the first image on the left in Figure 6.28. Repeat the process to find the remaining points of intersection. The results are shown in the last two images in Figure 6.28. Intersection K=-£ / v=-H Intersection K=0' / Y=0 i y Intersection K=2' i V=B Figure 6.28: Finding the points of intersection. Thus, the graphical solutions are x = —2, x = 0, and x = 2. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with x and y, respectively (see Figure 6.29). • Place your WINDOW parameters at the end of each axis (see Figure 6.29). • Label the graph with its equation (see Figure 6.29). 6.5. FACTORING SPECIAL FORMS 451 • Drop dashed vertical lines through each x-intercept. Shade and label the a;- values of the points where the dashed vertical line crosses the a;- axis. These are the solutions of the equation x 3 = 4x (see Figure 6.29). Figure 6.29: Reporting your graphical solution on your homework. Finally, note that the graphical solutions x = —2, x = 0, and x = 2 match our algebraic solutions exactly. Answer: —4, 0, 4 Intersection K=-4 V="fiH Intersection K=0 Y=0 ^ Intersection V=fiH □ 452 CHAPTER 6. FACTORING f> it, t* Exercises ■** •** ■** In Exercises 1-8, expand each of the given expressions. 1. (8r - 3t) 2 5. (s 3 - 9) 2 2. (6a + c) 2 6. (w 3 + 7) 2 3. (4a + 76) 2 7. (s 2 +6i 2 ) 2 4. (4s + t) 2 8. (7u 2 -2w 2 ) 2 In Exercises 9-28, factor each of the given expressions. 9. 25s 2 + 60s* + 36£ 2 19. 49r 6 + 112r 3 + 64 10. 9m 2 + 24uv + 16v 2 20. a e - 16a 3 + 64 11. 36m 2 - QOvw + 25w 2 21. 5s 3 £ - 20s 2 f 2 + 20st 3 12. 496 2 - 426c + 9c 2 22. 12r 3 t - 12r 2 i 2 + 3rt 3 13. a 4 + 18a 2 6 2 + 816 4 23. 8a 3 c + 8a 2 c 2 + 2ac 3 14. 64m 4 - U4u 2 w 2 + 81w 4 24. 18a; 3 z - 60a; 2 z 2 + 50xz 3 15. 49s 4 - 28s 2 i 2 + At 4 25. -486 3 + 1206 2 - 756 16. 4a 4 - 12a 2 c 2 + 9c 4 26. -45c 3 + 120c 2 - 80c 17. 496 6 - 112b 3 + 64 27. -5m 5 - 30u 4 - 45m 3 18. 25ir 6 - lOz 3 + 1 28. -12z 5 - 36z 4 - 27z 3 In Exercises 29-36, expand each of the given expressions. 29. (21c + 16)(21c - 16) 33. (3j/ 4 + 23z 4 )(3y 4 - 23z 4 ) 30. (19i + 7)(19t - 7) 34. (5x 3 + z 3 ){5x 3 - z 3 ) 31. (5a; + 19z)(5a: - 19z) 35. (8r 5 + 19s 5 )(8r 5 - 19s 5 ) 32. (11m + 5w)(11m - 5w) 36. (3m 3 + 16w 3 )(3m 3 - 16w 3 ) 6.5. FACTORING SPECIAL FORMS 453 In Exercises 37-60, factor each of the given expressions. 37. 361a; 2 - 529 49. 72y 5 - 242y 3 38. 96 2 - 25 50. 75y 5 - U7y 3 39. 16v 2 - 169 51. 1444a 3 o - 324a6 3 40. 81r 2 - 169 52. 126 3 c - 18756c 3 41. 169a; 2 - 576y 2 53. 576a; 3 z - 1156o;;z 3 42. 100y 2 - 81z 2 54. 192w 3 v - 507uv 3 43. 529r 2 - 289s 2 55. 576i 4 - 4£ 2 44. 49a 2 - 1446 2 56. 4z 5 - 256z 3 45. 49r 6 - 256£ 6 57. 81a; 4 - 256 46. 361a; 10 - 484z 10 58. 81a; 4 - 1 47. 36u 10 - 25w 10 59. 81a; 4 - 16 48. a 6 - 81c 6 60. a; 4 - 1 In Exercises 61-68, factor each of the given expressions completely. 61. z 3 + z 2 -9z-9 62. 3m 3 + u 2 -48m- 16 63. a; 3 - 2a; 2 y - xy 2 + 2y 3 64. a; 3 + 2a; 2 z - 4a;z 2 - 8z 3 In Exercises 69-80, solve each of the given equations for x. 69. 2a; 3 + 7a; 2 = 72a; + 252 75. 16a; 2 = 169 70. 2a; 3 + 7a; 2 = 32a; +112 76. 289a; 2 = 4 71. a; 3 + 5a; 2 = 64a; + 320 77. 9a; 2 = 25 72. a; 3 + 4a; 2 = 49a; + 196 78. 144a; 2 = 121 73. 144a; 2 + 121 = 264x 79. 256a; 2 + 361 = -608a; 74. 361a; 2 + 529 = 874a; 80. 16a; 2 + 289 = -136a; 65. r 3 - 3r 2 t - 25rt 2 + 75t 3 66. 26 3 - 36 2 c - 506c 2 + 75c 3 67. 2a; 3 + a; 2 - 32a; - 16 68. r 3 - 2r 2 - r + 2 454 CHAPTER 6. FACTORING In Exercises 81-84, perform each of the following tasks: i) Use a strictly algebraic technique to solve the given equation. ii) Use the 5:intersect utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown in Example 12. 81. x 3 = x 83. Ax 3 = x 82. x 3 = 9x 84. 9x 3 = x &»■ j*- £*■ Answers •*$ •** *** 1. 64r 2 -48rt + 9i 2 35. 64r 10 - 361s 10 3. 16a 2 + 56a6 + 496 2 37. (19a: + 23)(19x - 23) 5. s 6 - 18s 3 + 81 39. (4w + 13)(4v-13) 7. s 4 + 12s 2 i 2 + 36i 4 41. 169x 2 - 576y 2 9. (5s + 6i) 2 43. 529r2 " 289s 2 11. (6t, - 5u,) 2 45. (7r 3 + 16i 3 )(7r 3 - 16£ 3 ) 13. (a 2 + 96 2 ) 2 47. (6m 5 + 5w 5 )(6m 5 - 5u> 5 ) 15. (7s 2 - 2i 2 ) 2 49 - 2y 3 (6y + H)(6» - 11) 17 (7fe3_ 8 )2 51. 4a6(19a + 96)(19a-96) 19. (7r 3 + 8) 2 53- 4a:2; ( 12a; + 17z)(12a: - 17z) 21.5st(s-2tr 55.4t 2 (m+l)(m-l) 23.2 ac( 2a + C ) 2 57. (9x 2 + 16)(3x + 4)0,-4) 25. -36(46-5) 2 29. 441c 2 - 256 31. 25ir 2 -361z 2 33. 9y 8 - 529z 8 59. (9x 2 + 4)(3x + 2)(3x - 2) 27.-5.3 (u + 3 ) 2 81.(* + 3)(z-3)(z + l) 63. (x + j/)(x-y)(x-2j/) 65. (r + U)(r - 5t)(r - 3t) 67. (x + 4)(x-4)(2x+l) 69. x = —6, 6, 71. x = -8, 8, -5 6.5. FACTORING SPECIAL FORMS 455 7 73. x = 11 12 75. z = 13 13 ~T'T 77. a; = 5 5 ~3' 3 79. CE = 19 ~16 81. a; = 0,-1,1 83. x = 0,-1/2,1/2 456 CHAPTER 6. FACTORING 6.6 Factoring Strategy When you are concentrating on factoring problems of a single type, after doing a few you tend to get into a rhythm, and the remainder of the exercises, because they are similar, seem to flow. However, when you encounter a mixture of factoring problems of different types, progress is harder. The goal of this section is to set up a strategy to follow when attacking a general factoring problem. If it hasn't already been done, it is helpful to arrange the terms of the given polynomial in some sort of order (descending or ascending). Then you want to apply the following guidelines. Factoring Strategy. These steps should be followed in the order that they appear. 1. Factor out the greatest common factor (GCF). 2. Look for a special form. a) If you have two perfect squares separated by a minus sign, use the difference of squares pattern to factor: a 2 -b 2 = {a + b)(a-b) b) If you have a trinomial whose first and last terms are perfect squares, you should suspect that you have a perfect square trinomial. Take the square roots of the first and last terms and factor as follows. a 2 + 2ab + b 2 = (a + b) 2 Be sure to check that the middle term is correct. 3. If you have a trinomial of the form ax 2 + bx + c, use the ac- method to factor. 4. If you have a four-term expression, try to factor by grouping. Once you've applied the above strategy to the given polynomial, it is quite possible that one of your resulting factors will factor further. Thus, we have the following rule. Factor completely. The factoring process is not complete until none of your remaining factors can be factored further. This is the meaning of the phrase, "factor completely." 6.6. FACTORING STRATEGY 457 Finally, a very good word of advice. Check your factoring by multiplying. Once you've factored the given polynomial completely, it is a very good practice to check your result. If you multiply to find the product of your factors, and get the original given polyno- mial as a result, then you know your factorization is correct. It's a bit more work to check your factorization, but it's worth the effort. It helps to eliminate errors and also helps to build a better understanding of the factoring process. Remember, factoring is "unmultiplying," so the more you multiply, the better you get at factoring. Let's see what can happen when you don't check your factorization! Warning! ! The following solution is incorrect! Factor: 2a; 4 + 8a; 2 . Solution: Factor out the GCF. 2a; 4 + 8x 2 -■ = 2x 2 (x 2 +4) = 2x 2 {x + 2) 2 Note that this student did not bother to check his factorization. Let's do that for him now. Check: Multiply to check. Remember, when squaring a binomial, there is a middle term. 2x 2 {x + 2) 2 = 2x 2 (x 2 + 4x + 4) = 2a; 4 + 8a; 3 + 8a; 2 This is not the same as the original polynomial 2a; 4 + 8a; 2 , so the student's factorization is incorrect. Had the student performed this check, he might have caught his error, provided of course, that he multiplies correctly during the check. The correct factorization follows. 2a; 4 + 8a; 2 = 2a; 2 (a; 2 +4) The sum of squares does not factor, so we are finished. Check: Multiply to check. 2a; 2 (a; 2 +4) = 2a; 4 + 8a; 2 This is the same as the original polynomial 2a; 4 + 8a; 2 , so this factorization is correct. 458 CHAPTER 6. FACTORING You Try It! Factor completely: -4a; 7 + 64a; 3 Answer: -4x 3 (a; 2 + 4)(x + 2){x - 2) EXAMPLE 1. Factor completely: -3a; 6 3a; 2 Solution: The first rule of factoring is "Factor out the GCF." The GCF of — 3x 6 and 3a: 2 is 3a; 2 , so we could factor out 3a; 2 . -3a; 6 + 3a; 2 = 3a: 2 (-a: 4 + 1) This is perfectly valid, but we don't like the fact that the second factor starts with —x. Let's factor out —3a; 2 instead. -3a; 6 + 3a; 2 3a; 2 (a; 4 -l) The second factor is the difference of two squares. Take the square roots, separating one pair with a plus sign, one pair with a minus sign. = -3x 2 (a; 2 + l)(a; 2 - 1) The sum of squares does not factor. But the last factor is the difference of two squares. Take the square roots, separating one pair with a plus sign, one pair with a minus sign. = -3x 2 (a; 2 + l)(a; + l)(a;- 1) Check: Multiply to check the result. -3a; 2 (a; 2 + l)(x + l)(x - 1) = -3a; 2 (a; 2 + l)(a; 2 - 1) = -3a: 2 (a: 4 -l) -3a; 6 + 3a; 2 The factorization checks. □ You Try It! Factor completely: 3a 2 6 4 + 12a 4 & 2 - 12a 3 £ EXAMPLE 2. Factor completely: a: 3 y + 9a;y 3 + 6a; 2 y 2 Solution: The first rule of factoring is "Factor out the GCF." The GCF of x 3 y, 9a:?/ 3 , and 6x 2 y 2 is xy, so we factor out xy. x y + 9xy + 6x y = xy(x +9y + 6xy) Let's order that second factor in descending powers of x. = xy(x 2 + 6xy + 9y 2 ) 6.6. FACTORING STRATEGY 459 The first and last terms of the trinomial factor are perfect squares. We suspect we have a perfect square trinomial, so we take the square roots of the first and last terms, check the middle term, and write: xy(x + 3j/) 5 Thus, x 3 y + 9xy 3 + 6x 2 y 2 = xy(x + 3y) 2 . Check: Multiply to check the result. xy(x + 3y) 2 = xy(x 2 + 6xy + 9y 2 ) 9xy 3 x 3 y + 6x 2 y 2 Except for the order, this result is the same as the given polynomial. The factorization checks. Answer: 3a 2 6 2 (2a a EXAMPLE 3. Factor completely: 2x 3 - 48a; + 20a; 2 Solution: In the last example, we recognized a need to rearrange our terms after we pulled out the GCF. This time, let's arrange our terms in descending powers of x right away. 2a; 3 - 48a; + 20a; 2 2x J 2Qx z - 48a; You Try It! Factor completely: 27a; 3 - 3a; 4 - 60a; 2 Now, let's factor out the GCF. 2a;(a; 2 + 10a; - 24) The last term of the trinomial factor is not a perfect square. Let's move to the ac-method to factor. The integer pair —2, 12 has a product equal to ac = —24 and a sum equal to b = 10. Because the coefficient of a; 2 is one, this is a "drop in place" situation. We drop our pair in place and write: Thus, 2x 3 = 2x(x- 2)(x + 12) 48a; + 20a; 2 = 2a:(a; - 2)(x + 12). Check: Multiply to check the result. We use the FOIL method shortcut and mental calculations to speed things up. 2a;(a; - 2)(a; + 12) = 2a;(a; 2 + 10a; - 24) = 2a; 3 + 20a; 2 - 48a; Except for the order, this result is the same as the given polynomial. The factorization checks. Answer: — 3a; 2 (a; — 4)(a; — 5) □ 460 CHAPTER 6. FACTORING You Try It! Factor completely: 8a; 2 + 14xy - 15y 2 EXAMPLE 4. Factor completely: 2a 2 - 13a6 246 2 Solution: There is no common factor we can factor out. We have a trinomial, but the first and last terms are not perfect squares, so let's apply the ac-method. Ignoring the variables for a moment, we need an integer pair whose product is ac = —48 and whose sum is —13. The integer pair 3, —16 comes to mind (if nothing comes to mind, start listing integer pairs). Break up the middle term into a sum of like terms using the integer pair 3, —16, then factor by grouping. 2a 2 - 13a6 - 246 2 = 2a 2 + 3a6 - 16a6 - 246 2 = a(2a + 36) - 86(2a + 36) = (a-86)(2a + 36) Thus, 2a 2 - 13a6 - 246 2 = (a - 86)(2a + 36). Check: Multiply to check the result. We use the FOIL method shortcut and mental calculations to speed things up. (a - 86)(2a + 36) = 2a 2 - 13a6 - 246 2 Answer: (2x + 5y)(Ax — 3y) This result is the same as the given polynomial. The factorization checks. □ You Try It! Factor completely: 36a; 3 + 60a; 2 + 9a; EXAMPLE 5. Factor completely: 30a; 4 + 38a; 3 - 20x 2 Solution: The first step is to factor out the GCF, which in this case is 2a; 2 . 30a; 4 + 38a; 3 - 20a; 2 = 2a; 2 (15a; 2 + 19a; - 10) The first and last terms of the trinomial factor are not perfect squares, so let's move again to the ac-method. Comparing 15a; 2 + 19a; — 10 with ax 2 +bx+c, note that ac = (15) (—10) = —150. We need an integer pair whose product is —150 and whose sum is 19. The integer pair —6 and 25 satisfies these requirements. Because a/1, this is not a "drop in place" situation, so we need to break up the middle term as a sum of like terms using the pair —6 and 25. = 2a; 2 (15a; 2 - 6a; + 25a; - 10) Factor by grouping. Factor 3a; out of the first two terms and 5 out of the third and fourth terms. 2a; 2 (3a;(5a;-2) + 5(5a;-2)) 6.6. FACTORING STRATEGY 461 Finally, factor out the common factor 5a; — 2. = 2a; 2 (3a; + 5)(5a;-2) Thus, 30a: 4 + 38a; 3 - 20a: 2 = 2a: 2 (3a; + 5) (5a; - 2). Check: Multiply to check the result. Use the FOIL method to first multiply the binomials. 2a; 2 (3a; + 5) (5a; - 2) = 2a; 2 (15a; 2 + 19a; - 10) Distribute the 2a; 2 . = 30a; 4 + 38a; 3 - 20a; 2 This result is the same as the given polynomial. The factorization checks. Answer: 3a;(6a: + l)(2x + 3) □ You Try It! EXAMPLE 6. Factor completely: 8a; 5 + 10a; 4 - 72a; 3 - 90a; 2 Solution: Each of the terms is divisible by 3a; 3 . Factor out 3a; 3 . 15a; 6 - 33a; 5 - 240a; 4 + 528a; 3 = 3a; 3 [5a; 3 - 11a; 2 - 80a; + 176] The second factor is a four-term expression. Factor by grouping. = 3a; 3 [a: 2 (5a; - 11) - 16 (5a; - 11)] = 3a; 3 (a; 2 - 16) (5s- 11) The factor a; 2 — 16 is a difference of two squares. Take the square roots, separate one pair with a plus, one pair with a minus. = 3a; 3 (a; + 4)(a;-4)(5a;- 11) Thus, 15a; 6 - 33a; 5 - 240a; 4 + 528a; 3 = 3x 3 (a; + 4) (a; - 4) (5a; - 11). Check: Multiply to check the result. 3a; 3 (a; + 4) (a; - 4) (5a; - 11) = 3x 3 (a; 2 - 16) (5a; - 11) = 3a; 3 (5a; 3 - 11a; 2 - 80a; + 176 = 15a; 6 - 33a; 5 - 240a; 4 + 528a; 3 This result is the same as the given polynomial. The factorization checks. Factor completely: 15a; 6 - 33a; 5 - 240a; 4 528a; 3 Answer: 2a; 2 (a;-3)(a; + 3)(4a; + 5) □ 462 CHAPTER 6. FACTORING Using the Calculator to Assist the ac-Method When using the ac-method to factor ax 2 + bx + c and ac is a very large number, then it can be difficult to find a pair whose product is ac and whose sum in b. For example, consider the trinomial: 12y 2 - lly-36 We need an integer pair whose product is ac = —432 and whose sum is 6 = —11. We begin listing integer pair possibilities, but the process quickly becomes daunting. 432 216 Note that the numbers in the second column are found by dividing ac = —432 by the number in the first column. We'll now use this fact and the TABLE feature on our calculator to pursue the desired integer pair. 1. Enter the expression -432/X into Yl in the Y= menu (see the first image in Figure 6.30). 2. Above the WINDOW button you'll see TBLSET. Use the 2nd key, then press the WINDOW button to access the menu shown in the second image of Figure 6.30. Set TblStart=l, ATbl=l, then highlight AUTO for both the independent and dependent variables. 3. Above the GRAPH button you'll see TABLE. Use the 2nd key, then press the GRAPH button to access the table shown in the third image in Figure 6.30. Use the up- and down-arrow keys to scroll through the contents of the table. Note that you can ignore most of the pairs, be- cause they are not both integers. Pay attention only when they are both integers. In this case, remember that you are searching for a pair whose sum is b = —11. Note that the pair 16, —27 shown in the third image of Figure 6.30 is the pair we seek. Ploti PlOtE plots WiB- -432/X -:Vi=U Wj = ^Vh= We = ^Vi= \V? = TABLE SETUP TblStart=l ^Tbl=l Indpnt: grog Rsk Depend: flsk X V1 Ff^^B -39.27 12 "3fi 13 -33.23 1H -30.B6 IE -2B.B IS -27 1? -25.11 X=ll Figure 6.30: Using the TABLE feature to assist the ac-method. 6.6. FACTORING STRATEGY 463 Now we can break the middle term of I2y 2 — lly — 36 into a sum of like terms using the ordered pair 16, — 27, then factor by grouping. • 2 -iiy- -36 = = 12y 2 + 162/- -27y- -36 = 4y(3y + 4)- -9(32/ + 4) = (4y- - 9)(3y + 4) Check: Use the FOIL method shortcut and mental calculations to multiply. (42/-9)(32/ + 4) = 122/ 2 -ll2/-36 The factorization checks. 3. 3x 7 z b - 363x 5 ;j 5 4. 5r 5 s 2 - 80r 3 s 2 5. 2w 7 - 162u 5 6. 405a; 4 - 320a; 2 464 CHAPTER 6. FACTORING f> t* t* Exercises ■** •** ■** In Exercises 1-12, factor each of the given polynomials completely. 1. 484y 4 z 2 - U4y 2 z 4 7. 3t> 8 - 1875w 4 2. 72s 4 i 4 - 242s 2 t 6 8. 3a 9 - 48a 5 9. 3x 6 - 300a; 4 10. 2y 5 - 18y 3 11. 1250u 7 w 3 - 2w 3 w 7 12. 48y 8 z 4 - 3y 4 z 8 In Exercises 13-24, factor each of the given polynomials completely. 13. 75a 6 - 210a 5 + 147a 4 19. 2^ 4 - 4z 3 + 2z 2 14. 245v 7 - 560u 6 + 320w 5 20. 2m 6 - 40w 5 + 200m 4 15. 180a 5 6 3 + 540a 4 6 4 + 405a 3 6 5 21. 324a; 4 + 360a; 3 + 100x 2 16. 192uV + 432uV + 243u 4 w 6 22. 986 4 + 846 3 + 186 2 17. 26 5 + 46 4 + 26 3 23. 756 4 c 5 - 2406 3 c 6 + 1926 2 c 7 18. 3v 6 + 30v 5 + 75u 4 24. 162a 5 c 4 - 180a 4 c 5 + 50a 3 c 6 In Exercises 25-36, factor each of the given polynomials completely. 25. 5a 5 + 5a 4 - 210a 3 31. 4a 6 + 64a 5 + 252a 4 26. 3y 5 - 9y 4 - 12y 3 32. 4x 4 + 64a; 3 + 252a; 2 27. 3y 6 - 39y 5 + 120w 4 33. 3z 4 + 33z 3 + 84z 2 28. 3y 7 - 27y 6 + 42y 5 34. 5a 6 + 65a 5 + 180a 4 29. 3z 4 + I2z 3 - I3hz 2 35. bz 7 - 75z 6 + 270z 5 30. 5a 4 - 40a 3 - 45a 2 36. 3y 4 - 27y 3 + 24y 2 6.6. FACTORING STRATEGY 465 In Exercises 37-48, factor each of the given polynomials completely. 37. 46 3 - 22b 2 + 306 43. 12a; 3 + 9a; 2 - 30a; 38. 46 6 - 22b 5 + 30b 4 44. 6v 4 + 2v 3 - 20v 2 39. 2u 4 w 5 - 3u 3 w 6 - 20u 2 w 7 45. 8m 6 + 34w 5 + 30u 4 40. 12x 5 z 2 + 9x 4 z 3 - 30x 3 z 4 46. 4a 4 + 29a 3 + 30a 2 41. 12x 4 y 5 + 50x 3 y 6 + 50x 2 y 7 47. 12a 4 c 4 - 35a 3 c 5 + 25a 2 c 6 42. 24s 4 t 3 + 62s 3 i 4 + 40s 2 i 5 48. 18x 6 z 5 - 39x 5 z 6 + 18x 4 z 7 In Exercises 49-56, factor each of the given polynomials completely. 49. 12y 5 + 15y 4 - 108y 3 - 135y 2 53. 72z 6 + 108z 5 - 2z 4 - 3z 3 50. 96 8 + 126 7 - 3246 6 - 4326 5 54. 216a; 7 + 324a; 6 - 6a; 5 - 9a; 4 51. 9a; 6 ^ 5 + 6a; 5 ;z 6 - lUx 4 z 7 - 96x 3 z s 55. 144a 6 c 3 + 360a 5 c 4 - 4a 4 c 5 - 10a 3 c 6 52. 12w 7 w 3 + 9u 6 w 4 - 432u 5 w 5 - 324w 4 w 6 56. 48a 8 c 4 + 32a 7 c 5 - 3a 6 c 6 - 2a 5 c 7 In Exercises 57-60, use your calculator to help factor each of the given trinomials. Follow the procedure outline in Using the Calculator to Assist the ac-Method on page 462. 57. 6a; 2 + 61a; + 120 59. 60a; 2 - 167a; + 72 58. 16a; 2 - 62a; - 45 60. 28a; 2 + x - 144 $»• 2** £*■ Answers ■*$ -as >*$ 1. 4y 2 z 2 (lly + 6z)(lly - 6z) 13. 3a 4 (5a - 7) 2 3. 3a; 5 z 5 (a; + ll)(a; - 11) 15 - 5a 3 6 3 (6a + 96) 2 17. 26 3 (6+l) 2 19. 2z 2 (z- l) 2 21. 4a; 2 (9a; + 5) 2 23. 36 2 c 5 (56-8c) 2 11. 2w 3 w 3 (25u 2 + w 2 )(5u + w){5u - w) 25. 5a 3 (a - 6)(a + 7) 5. 2u 5 {u + 9)(u-9) 7. 3v 4 (v 2 + 25){v + 5)(v-5) 9. 3a; 4 (x + 10) {x - 10) oq q^s/^ o„\2 466 27. 3y 4 (y-8)G/-5) 29. 3z 2 (;z-5)0 + 9) 31. 4a 4 (a + 9)(a + 7) 33. 3z 2 (x + 7)(z + 4) 35. 5z 5 (z-6)0-9) 37. 26(26 -5)(6- 3) 39. u 2 w 5 (u - 4w)(2w + 5w) 41. 2x 2 y 5 {3x + by)(2x + by) 43. 3x(4x - 5)(x + 2) CHAPTER 6. FACTORING 45. 2u 4 (4u + 5)(w + 3) 47. a 2 c 4 (4a-5c)(3o-5c) 49. 3y 2 (y + 3)(y-3)(4 2 / + 5) 51. 3x 3 z 5 (x + 4z)(x - 4z)(3x + 2z) 53. z 3 (6z + l)(6z- l)(2z + 3) 55. 2a 3 c 3 (6a + c)(6a - c)(2a + 5c) 57. (2a! + 15)(3a; + 8) 59. (15x-8)(4a;-9) 6. 7. APPLICATIONS OF FACTORING 467 6.7 Applications of Factoring In this section we will solve applications whose solutions involve factoring. Let's begin. EXAMPLE 1. A rectangular canvas picture measures 12 inches by 16 inches. The canvas is mounted inside a frame of uniform width, increasing the total area covered by both canvas and frame to 396 square inches. Find the uniform width of the frame. Solution: We follow the Requirements for Word Problem Solutions. 1. Set up a variable dictionary. A carefully labeled figure will help us main- tain our focus. We'll let x represent the uniform width of the frame. You Try It! A rectangular canvas picture measures 7 inches by 11 inches. The canvas is mounted inside a frame of uniform width, increasing the total area covered by both canvas and frame to 117 square inches. Find the uniform width of the frame. 16 + 2a: X 16 X 12 12 X 16 X Yl + 2x 16 + 2a: 2. Set up an equation. If the inner rectangular dimensions are 16 inches by 12 inches, adding a frame of uniform width x makes the dimensions of the frame plus canvas 16 + 2x inches by 12 + 2x inches. The total area is found by multiplying these outer dimensions, A = (16 + 2x)(12 + 2x). If the total area is A = 396 square inches, then we have the following equation. (16 + 2x)(12 + 2a;) = 396 3. Solve the equation. We start by expanding the right-hand side of the equation. (16 + 2a;)(12 + 2a;) = 396 192 + 56z + Ax 2 = 396 The resulting equation is nonlinear. Make one side zero. 4oT + 56a; - 204 = 468 CHAPTER 6. FACTORING Answer: 1 inch We could factor out 4 on the left-hand side, but since there is a zero on the right-hand side of the equation, it's a bit easier to simply divide both sides by 4. Note how we divide each term on the left-hand side by the number 4. x 2 + 14a; -51 = We need an integer pair whose product is ac = —51 and whose sum is b = 14. The integer pair —3, 17 comes to mind. Since the coefficient of x 2 is one, we can simply "drop in place" our ordered pair. (z-3)(x + 17) = Thus, the solutions are x = 3 and x = —17. 4. Answer the question. The uniform width of the frame cannot be —17 inches, so we eliminate this solution from consideration. Hence, the uni- form width of the frame is 3 inches. 5. Look back. If the uniform width of the frame is 3 inches, then the final dimensions will look like the following. 22 3 16 3 t > 12 16 12 3 < > 3 18 22 Thus, the combined area of the frame plus canvas is A = (18) (22), or A = 396 square inches, the area given in the problem statement. Our solution is correct. □ You Try It! A projectile's height (in feet) is given by the equation y= _i6t 2 + 144i + 576, where time t is measured in seconds. How much time passes before the projectile hits the ground? EXAMPLE 2. A projectile is fired at an angle into the air from atop a cliff overlooking the ocean. The projectile's distance (in feet) from the base of the cliff is give by the equation x = not, (6.1) 6. 7. APPLICATIONS OF FACTORING 469 and the projectile's height above sea level (in feet) is given by the equation y = _i6i 2 + 288i + 640, (6.2) where t is the amount of time (in seconds) that has passed since the projectile's release. a) How much time passes before the projectile splashes into the ocean? b) At that time, how far is the projectile from the base of the cliff? Solution: We follow the Requirements for Word Problem Solutions. 1. Set up a variable dictionary. The variables are already set up for us. x = Distance from base of the cliff (in feet). y = Height above sea level (in feet). t = Time since projectile's release (in seconds). 2. Set up an equation. The equations are already set up (see equation 6.1 and equation 6.2). 3. Solve the equation. When the projectile splashes into the ocean, its height above sea level at that moment is y = feet. Substitute for y in equation 6.2 and solve the resulting equation for t. y = _i6t 2 + 288i + 640 = -I6i 2 + 288^ + 640 We could factor out —16, but since the left-hand side of this equation is zero, it's a bit easier to divide both sides by —16. Note how we divide each term on the right-hand side by —16. = t 2 - 18* - 40 We need a pair of integers so that their product is ac = —40 and their sum is —18. The integer pair 2, —20 comes to mind. Since the coefficient of t 2 is one, we can "drop in place" our integer pair. = (£ + 2)(£-20) Hence, the solutions are t = — 2 and t = 20. 4. Answer the question. In answering question (a), the solution t = — 2 seconds makes no sense. Thus, the projectile splashes into the ocean at t = 20 seconds. 470 CHAPTER 6. FACTORING Answer: 12 seconds In addressing question (b), to find the projectile's distance from the base of the cliff at this moment, substitute t = 20 in equation 6.1. x = 120* x = 120(20) x = 2400 Hence, at the moment the projectile splashes into the ocean, it is 2,400 feet from the base of the cliff. 5. Look back. The best we can do here is check our solution * = 20 in equation 6.2. y = _i6* 2 + 288* + 640 y = -16(20) 2 + 288(20) + 640 y = -6400 + 5760 + 640 y = Indeed, at * = 20, the projectile does splash into the ocean. □ You Try It! The product of two consecutive positive odd integers is 483. Find the integers. EXAMPLE 3. The product of two consecutive even integers is 728. Find the integers. Solution: We follow the Requirements for Word Problem Solutions. 1. Set up a variable dictionary. Let k represent the first even integer. Then k + 2 represents the next consecutive even integer. 2. Set up an equation. The product of the integers is 728. Hence, we have the following equation. k(k + 2) = 728 3. Solve the equation. Expand the left-hand side of the equation. k 2 + 2k = 728 The equation is nonlinear. Make one side zero. k 2 + 2k - 728 = See Using the Calculator to Assist the ac-Method on page 462. We need an integer pair whose product is ac = —728 and whose sum is b = 2. Enter -728/X in Yl, then set up the table (see Figure 6.31). 6. 7. APPLICATIONS OF FACTORING 471 Plotl FlotE Plots WiB-728^X \Yi = We = TRBLE SETUP TblStart=l aTbl=l Indpnt: rang Rsk Depend: I3KSI Rsk X Vi 21 -30.33 25 "29.12 26 -SB 2? "Efi.Sfi 2B "Efi b "25.1 -EH.E? X=3@ Figure 6.31: Load ac/X , or -728/X into Yl in the Y= menu. Use the up- and down-arrow keys to scroll. Note that 28, —26 is the desired pair. Because the coefficient of k 2 is one, we can "drop in place" the ordered pair. = (fc + 28)(fc-26) Hence, the solutions are k = —28 and k = 26. 4. Answer the question. If k = —28, the next consecutive even integer is k + 2 = —26. Secondly, if k = 26, the next consecutive even integer is fc + 2 = 28. 5. Lookback. Our first pair is —28 and —26. They have the required product of 728. Our second pair is 26 and 28. Their product is also 728. Both solutions check! Answer: 21 and 23 □ EXAMPLE 4. A rectangle has perimeter 54 feet and area 180 square feet. Find the dimensions of the rectangle. Solution: We follow the Requirements for Word Problem Solutions. You Try It! A rectangle has perimeter 62 feet and area 234 square feet. Find the dimensions of the rectangle. 1. Set up a variable dictionary. A sketch will help us keep our focus. Let L represent the length of the rectangle and let W represent its width. W w 472 CHAPTER 6. FACTORING 2. Set up an equation. The perimeter is 54 feet, so 2W+2L = 54, or dividing both sides by 2: W + L = 27 (6.3) The area is 180 square feet, so: LW = 180 (6.4) 3. Solve the equation. The system of equations (equations 6.3 and 6.4) can be solved using the substitution method. First, solve equation 6.3 for W: W = 27-L (6.5) Substitute equation 6.5 in equation 6.4, expand, then make one side zero. 180 L(27-L) 27 X - X 2 180 = L 1 - 27L + 180 The integer pair —12,-15 has product ac = 180 and sum b = —27. Moreover, the coefficient of X 2 is one, so we can "drop in place" our integer pair. = (X- 12) (X- 15) Hence, the solutions are X = 12 and X = 15. 4. Answer the question. Two possibilities for the width. Substitute X = 12 in (6.5). Substitute L = 15 in (6.5). W = 27-L W = 27- 12 W = 15 W = 27-L W = 27-15 W = 12 Both answers give the same 15 by 12 foot rectangle, but we usually think of the term "length" as the longer side of the rectangle. So let's go with the length is L = 15 feet and the width is W = 12 feet. 5. Look back. Let's add L = 15 feet and W = 12 feet to a diagram. 15 ft 12 ft 12 ft 15 ft 6. 7. APPLICATIONS OF FACTORING 473 If we add the dimensions around the rectangle, the perimeter is P = 15 + 12 + 15 + 12, or P = 54 feet, the perimeter required in the problem statement. Next, if we multiply the dimensions, then A = (15)(12), or A = 180 square feet, the area required in the problem statement. Our Answer- length = 18 feet solution is correct! and width = 13 feet □ 474 CHAPTER 6. FACTORING f»- n- fa- Exercises ■*1 ■*'» .*5 1. A rectangular canvas picture measures 14 inches by 36 inches. The canvas is mounted inside a frame of uniform width, increasing the total area covered by both canvas and frame to 720 square inches. Find the uniform width of the frame. 2. A rectangular canvas picture measures 10 inches by 32 inches. The canvas is mounted inside a frame of uniform width, increasing the total area covered by both canvas and frame to 504 square inches. Find the uniform width of the frame. 3. A projectile is fired at an angle into the air from atop a cliff overlooking the ocean. The projectile's distance (in feet) from the base of the cliff is give by the equation x = 180£, and the projectile's height above sea level (in feet) is given by the equation y -16£ 2 + 352^ + 1664, where t is the amount of time (in seconds) that has passed since the projectile's re- lease. How much time passes before the projectile splashes into the ocean? At that time, how far is the projectile from the base of the cliff? 4. A projectile is fired at an angle into the air from atop a cliff overlooking the ocean. The projectile's distance (in feet) from the base of the cliff is give by the equation x = UOt, and the projectile's height above sea level (in feet) is given by the equation where t is the amount of time (in seconds) that has passed since the projectile's re- lease. How much time passes before the projectile splashes into the ocean? At that time, how far is the projectile from the base of the cliff? 5. The product of two consecutive even inte- gers is 624. Find the integers. 6. The product of two consecutive even inte- gers is 528. Find the integers. 7. The product of two consecutive positive integers is 552. Find the integers. 8. The product of two consecutive positive integers is 756. Find the integers. 9. The product of two consecutive odd inte- gers is 483. Find the integers. 10. The product of two consecutive odd inte- gers is 783. Find the integers. 11. A rectangle has perimeter 42 feet and area 104 square feet. Find the dimensions of the rectangle. 12. A rectangle has perimeter 32 feet and area 55 square feet. Find the dimensions of the rectangle. 13. The radius of the outer circle is one inch longer than twice the radius of the inner circle. V -16r + 288* + 1408, If the area of the shaded region is 407T square inches, what is the length of the inner radius? 6. 7. APPLICATIONS OF FACTORING 475 14. The radius of the outer circle is two inches longer than three times the radius of the inner circle. 15. If the area of the shaded region is 1807T square inches, what is the length of the inner radius? You have two positive numbers. The sec- ond number is three more than two times the first number. The difference of their squares is 144. Find both positive num- bers. You have two positive numbers. The sec- ond number is two more than three times the first number. The difference of their squares is 60. Find both positive num- bers. 17. Two numbers differ by 5. The sum of their squares is 97. Find the two numbers. 16. 18. Two numbers differ by 6. The sum of their squares is 146. Find the two numbers. 19. The length of a rectangle is three feet longer than six times its width. If the area of the rectangle is 165 square feet, what is the width of the rectangle? 20. The length of a rectangle is three feet longer than nine times its width. If the area of the rectangle is 90 square feet, what is the width of the rectangle? 21. The ratio of the width to the length of a given rectangle is 2 to 3, or 2/3. If the width and length are both increased by 4 inches, the area of the resulting rectangle is 80 square inches. Find the width and length of the original rectangle. 22. The ratio of the width to the length of a given rectangle is 3 to 4, or 3/4. If the width is increased by 3 inches and the length is increased by 6 inches, the area of the resulting rectangle is 126 square inches. Find the width and length of the original rectangle. j*- j*- j*- Answers •*$ **s •** 1. 2 inches 3. 26 seconds, 4,680 feet 5. -26 and -24, and 24 and 26 7. 23, 24 9. -23 and -21, and 21 and 23 11. 8 feet by 13 feet 13. 3 inches 15. 5 and 13 17. 4 and 9, and -4 and -9 19. 5 feet 21. 4 inches by 6 inches Chapter 7 Rational Expressions Our ability to communicate and record ideas is enhanced by concise mathe- matical language and notation. More importantly, our ability to conceive new ideas is broadened by notation that the mind can effectively use to organize and recognize patterns. Exponential notation is a simple example of such nota- tion that expands our thinking. The modern notation we use started seriously in 1636 and 1637 with James Humes and Rene Descartes, after centuries of mathematicians flirting with various approaches. Descartes may have been the first to use today's radical symbol, combining the check symbol with the bar over the top in 1637. Scientific notation, as we define it today, is a relatively new notation, first appearing in the mid-1900's. One may wonder where the next significant notational breakthrough will originate. Maybe it will be your contribution. 477 478 CHAPTER 7. RATIONAL EXPRESSIONS 7.1 Negative Exponents We begin with a seemingly silly but powerful definition on what it means to raise a number to a power of — 1 . Raising to a Power of —1. To raise an object to a power of —1, simply invert the object (turn it upside down). 9 A v 6 More formally, inverting a number is known as taking its reciprocal. You Try It! Simplify: ( - EXAMPLE 1. Simplify each of the following expressions: -1 / o\ -i a) 4- b) Solution: In each case, we simply invert the given number. a) A- 1 1 b) 2\ X 3 Solution: In each case, we simply invert the given number, a) 4- = i b) (-5)- 1 1 2\ 1 3 d) 3V 1 5 Answer: — 7 □ 7.1. NEGATIVE EXPONENTS 479 You might be asking "Why does raising to the power of minus one invert the number?" To answer this question, recall the product of a number and its reciprocal is one. For example, 4-i = l. (7.1) Next, consider what happens when we multiply 4 1 and 4 _1 . If we apply the usual law of exponents (assuming they work for both positive and negative exponents), we would add the exponents (1 + (—1) = 0). 4 1 -4~ 1 =4° (7.2) However, because 4 1 = 4 and 4° = 1, this last equation is equivalent to: 4-4 _1 = l (7.3) When you compare Equation 7.1 and 7.3, it is clear that 4 _1 and 1/4 are both reciprocals of the number 4. Because reciprocals are unique, 4 _1 = 1/4. In similar fashion, one can discover the meaning of a~ n . Start with the fact that multiplying reciprocals yields an answer of one. a n • 4 = 1 (7.4) If we multiply a n and a - ™, we add the exponents as follows. a n ■ a~ n = a Providing a ^ 0, then a = 1, so we can write a n -a- n = l (7.5) Comparing Equations 7.4 and 7.5, we note that both l/o n and a~" are recip- rocals of a n . Because every number has a unique reciprocal, a~ n and 1/a" are equal. Raising to a negative integer. Provided o/:0, a™ 480 CHAPTER 7. RATIONAL EXPRESSIONS You Try It! Simplify: 3 Answer: EXAMPLE 2. Simplify each of the following expressions: >-3 b) (-5)" (-4) -3 Solution: In each example, we use the property a n = l/a n to simplify the given expression. a) 2- ' ' l l 2 3 1 b) (-5)" (-5)2 1 25 (-4)" (-4)3 1 ~64 In Raising to a Negative Integer on page 482, we'll address how you can per- form each of the above computations mentally. □ Laws of Exponents In the arguments demonstrating that 4 _1 = 1/4 and a~ n = 1/a™, we appealed to one of the laws of exponents learned in Chapter 5, Section 5. Fortunately, the laws of exponents work exactly the same whether the exponents are positive or negative integers. Laws of Exponents. If m and n are integers, then: 1. a m a n = a m+n a m 2. = a m - n a n 3. ( a m ) n = a mn 4. (ab) n = a n b n ■- id' b n You Try It! Simplify: t s ■ V EXAMPLE 3. Simplify each of the following expressions: b) 2- 2 -2" 3 a) y 5 y 7 x~ 4 x 6 Solution: In each case, we use the first law of exponents (a m a n = a m+n ). Be- cause we are multiplying like bases, we repeat the base and add the exponents. 7.1. NEGATIVE EXPONENTS 481 = y~ 2 b) 2- 2 • 2- 3 = 2- 2 +(- 3 > c) X -V = .*- 4 + 6 Answer: £ 4 D EXAMPLE 4. Simplify each of the following expressions: b) 3 5 Solution: In each case, we use the second law of exponents (a m /a n = a m ~ n ). Because we are dividing like bases, we repeat the base and subtract the expo- nents. Recall that subtraction means "add the opposite." .4-7 X 4+(-7) o-4 -4-5 -4+(-5) y-5 -3-(-5) y -3+5 You Try It! y Simplify: — - y-2 Answer: y D EXAMPLE 5. Simplify each of the following expressions: a) (5- 2 ) 3 b) (a- 3 )- 4 c) (w 2 )" 7 Solution: In each case, we are using the third law of exponents ((a m ) n = a mn ). Because we are raising a power to another power, we repeat the base and multiply the exponents. a) (5- 2 ) 3 =5(- 2 >( 3 > b) (a" 3 ) 3\-4 j(-3)(-4) ,12 C) («,') 2\-7 ,(2)(-7) W -14 You Try It! Simplify: (z ) 5\-2 Answer: a 482 CHAPTER 7. RATIONAL EXPRESSIONS You Try It! Simplify: ( - Raising to a Negative Integer We know what happens when you raise a number to — 1 , you invert the number or turn it upside down. But what happens when you raise a number to a negative integer other than negative one? As an example, consider the expression 3~ 2 . Using the third law of expo- nents ((a m )" = a" m ), we can write this expression in two equivalent forms. 1. Note that 3 -2 is equivalent to (3 2 )" 1 . They are equivalent because the third law of exponents instructs us to multiply the exponents when raising a power to another power. Finally, note that to evaluate (S 2 )^ 1 , we first square, then invert the result. (3 2 ) 9- 1 1 9 Repeat base and multiply exponents. Simplify: 3 = Simplify: 9~ 9. 1/9. Note that 3~ 2 is also equivalent to (3 -1 ) 2 . They are equivalent because the third law of exponents instructs us to multiply the exponents when raising a power to another power. Finally, note that to evaluate (3 -1 ) 2 , we first invert, then square the result. (3 -1\2 Repeat base and multiply exponents. Simplify: 3" 1 = 1/3. Simplify: (1/3) 2 = 1/9. Using either technique, 3 -2 = 1/9. You can either square and invert, or you can invert and square. In each case, the 2 means "square" and the minus sign means "invert," and this example shows that it doesn't matter which you do first. EXAMPLE 6. Simplify each of the following expressions: -2 / r,\ -3 a) 5~ 3 b)(- _ 4 )-2 Solution: a) We'll cube then invert 5- 3 = (5 3 )- 1 125 -1 1 125 d) Repeat base, multiply exponents. Simplify: 5 3 = 125. Invert: 125" 1/125. 7.1. NEGATIVE EXPONENTS 483 Note that the three means "cube" and the minus sign means "invert," so it is possible to do all of this work mentally: cube 5 to get 125, then invert to get 1/125. b) We'll square then invert. (— 4)~ = ((—4) )~ Repeat base, multiply exponents. = 16" 1 Simplify: (-4) 2 = 16. Invert: 16" 1 = 1/16. 1 16 Note that the two means "square" and the minus sign means "invert," so it is possible to do all of this work mentally: square —4 to get 16, then invert to get 1/16. c) Again, we'll square then invert. 3\ //3 X 9 25 25 ~9~ Repeat base, multiply exponents. Simplify: (3/5) 2 = 9/25. Invert: (9/25)" 1 = 25/9. Note that the two means "square" and the minus sign means "invert," so it is possible to do all of this work mentally: square 3/5 to get 9/25, then invert to get 25/9. d) This time we'll cube then invert. 2\ // 2 X 3/ \ V 3. 27 27 Repeat base, multiply exponents. Simplify: (-2/3) 3 = -8/27. Invert: (-8/27)" 1 = -27/8. Note that the three means "cube" and the minus sign means "invert," so it is possible to do all of this work mentally: cube —2/3 to get —8/27, then invert to get -27/8. 64 Answer: 125 □ 484 CHAPTER 7. RATIONAL EXPRESSIONS Applying the Laws of Exponents In this section we'll simplify a few more complicated expressions using the laws of exponents. You Try It! Simplify: (-5x s y- 2 )(-2x- 6 y- 1 ) EXAMPLE 7. Simplify: (2x- 2 y 3 ){-3x 5 y- 6 ) Solution: All the operators involved are multiplication, so the commutative and associative properties of multiplication allow us to change the order and grouping. We'll show this regrouping here, but this step can be done mentally. ^arVX-SxV 6 ) = [(2)(-3)](a:- 2 x 5 )(y 3 y- 6 ) When multiplying, we repeat the base and add the exponents. = _6x- 2 +V+(- 6 ) = -6x 3 2T 3 In the solution above, we've probably shown way too much work. It's far easier to perform all of these steps mentally, multiplying the 2 and the —3, then repeating bases and adding exponents, as in: Answer: lOx y 2, ,-3 „3„,-3 (2x- 2 y i )(-3x b y" b ) = -Qx 6 y □ You Try It! Simplify: _ (4x z y°) fix v EXAMPLE 8. Simplify: „ _ 9x 6 y 2 Solution: The simplest approach is to first write the expression as a product. 6x~ 2 y 5 6 x~ 2 y 5 9x 3 y~ 2 9 x 3 y~ 2 Reduce 6/9 to lowest terms. Because we are dividing like bases, we repeat the base and subtract the exponents. x -2- 3j/ 5-(-2) a ,-2+(-3) y 5+2 X~V 7.1. NEGATIVE EXPONENTS 485 In the solution above, we've probably shown way too much work. It's far easier to imagine writing the expression as a product, reducing 6/9, then repeating bases and subtracting exponents, as in: 2„,5 6x y 9x 3 y~ f*-y Answer: — x 5 y 6 2 y U You Try It! -2„,4\-3 EXAMPLE 9. Simplify: (2arV) Solution: The fourth law of exponents {{ab) n = a n b n ) says that when you raise a product to a power, you must raise each factor to that power. So we begin by raising each factor to the minus three power. Simplify: (3x 4 j/ To raise two to the minus three, we must cube two and invert: 2 -3 = 1/8. Secondly, raising a power to a power requires that we repeat the base and multiply exponents. l a; (-2)(-3) y (4)(-3) ;x V -12 In the solution above, we've probably shown way too much work. It's far easier raise each factor to the minus three mentally: 2 -3 = 1/8, then multiply each exponent on the remaining factors by —3, as in (2z-V) -x 6 y- 12 Answer: -x~ 8 y 6 a Clearing Negative Exponents Often, we're asked to provide a final answer that is free of negative exponents. It is common to hear the instruction "no negative exponents in the final an- swer." Let's explore a couple of techniques that allow us to clear our answer of negative exponents. 486 CHAPTER 7. RATIONAL EXPRESSIONS You Try It! Simplify the expression x~ 2 so that the resulting equivalent expression contains no negative exponents. EXAMPLE 10. Consider the expression: y-3 Simplify so that the resulting equivalent expression contains no negative expo- nents. Solution: Raising y to the —3 means we have to cube and invert, so y~ 3 = 1/y 3 . 1 2 X z X Answer: y 5 x 2 y To divide x 2 by 1/y 3 , we invert and multiply. „2 . x 2 y 3 1 I Alternate approach: An alternate approach takes advantage of the laws of exponents. We begin by multiplying numerator and denominator by y 3 . x 2 x 2 y 3 y-3 y-3 y3 x 2 y 3 x 2 y 3 In the last step, note how we used the fact that y = 1. a You Try It! Simplify the expression x~ 3 y 2 so that the resulting equivalent expression contains no negative exponents. EXAMPLE 11. Consider the expression: 2x 2 y~ 2 z 3 Simplify so that the resulting equivalent expression contains no negative expo- nents. 7.1. NEGATIVE EXPONENTS 487 Solution: Again, we can remove all the negative exponents by taking recip- rocals. In this case y~ 2 = 1/y 2 (square and invert). 2 l 2x 2 y~ 2 y 2 2x 2 yl To divide 2x jy by z , we invert and multiply. „2 yl -z 3 2x 2 1 V 2 ~ 2x 2 y 2 z 3 Alternate approach: An alternate approach again takes advantage of the laws of exponents. We begin by multiplying numerator and denominator by y 2 - 2x 2 y~ -2 2x 2 y~ 2 v 2 z 3 z 3 2x 2 y° y 2 z 3 2x 2 y2 z 3 y 2 2 A V z In the last step, note how we used the fact that y = 1. Answer: — 3x A D 488 CHAPTER 7. RATIONAL EXPRESSIONS f> t* t* Exercises ■** -** ■** In Exercises 1-8, simplify the given expression. '•or 4. nr / 3\ _1 5. (IS)" 1 2 h) 6. (-11)- 1 ( 8\ _1 7. (16)- 1 3 b) 8. (7)- 1 In Exercises 9-16, simplify the given expression. 9. a~ 9 a 3 13. 2 9 . 2 -4 10. :r~ 5 x~ 5 14. 2 2 -2- 7 11. 6- 9 6 8 15. 9- 6 -9" 5 12. v- 7 u- 2 16. 9 7 -9~ 5 In Exercises 17-24, simplify the given expression. 2 6 17. 2~8 18. 6 8 6- 1 19. z- 1 z 9 20. —4 M> 3 21. _9 W w' 22. r 5 r -l 23. 7- 3 7- 1 24. 6 -8 6 6 In Exercises 25-32, simplify the given expression. 25. (r 1 ) 4 27. (6- 6 ) 7 26. (a s y 7 28. (2- 7 )" 7 7.1. NEGATIVE EXPONENTS 489 29. z" 30. (c 6 Y 31. (3- 2 ) : 32. (8- 1 )* In Exercises 37-36, simplify the given expression. 33. 4 3 34. 5~ 2 35. 2~ 4 36. (-3)" 38. 39. 37. 40. In Exercises 41-56, simplify the given expression. 41. (4u- 6 v~ 9 ) (5u 8 V 8 ) 42. (6a- 9 c- 6 ) (-8a 8 c 5 ) 43. (6x- 6 y- 5 ) (-4x 4 y- 44. (5v- 3 w- s ) (8v- 9 w 5 45. — 6x 7 z 9 <±x~ 9 z~ 2 46. 2u~ 2 v 6 6u 2 v 1 47. -6a 9 c 6 -4a" 5 c- 7 48. — 4u~ 4 u; 4 49. 2v- 2 w 4 y'° 50. 3.s- 6 i 5 )" 4 51. 3x-V) 4 52. -4 6- 8 c- 4 ) 3 53. 2x e z- 7 ) 5 54. '-4v A w~ 9 ) 3 55. 2a- 4 C 8 )" 4 56. ll^c- 1 ) -2 In Exercises 57-76, clear all negative exponents from the given expression. 57. 58. 59. x 5 y~ -2 z 3 4 - -9 z 7 Q — r s 2 p 60. 61. 62. y 8 z 5 y- 4 z 3 490 CHAPTER 7. RATIONAL EXPRESSIONS 63. — 1 —= 70 64. a 7 6- 8 c 6 65. (7a;- 1 )(-7a;- 1 ) 66. (3a- 8 )(-7a- 7 ) 67. (8a- 8 )(7a- 7 ) 68. (-7u 3 )(-8m- 6 ) fiq 4a;" 9 cS.r :i 71 72 -6i 9 6c 2 -4c 7 6v~ 9 8v- 4 73. (-3.s 9 )- 4 74. (-3s 8 )- 4 75. (2y 4 )- 5 76. (2w 4 )- 5 j*- j*- j*- Answers •*$ *** •** 1. 7 27. 6 7.1 16 13. 2 5 15. 9- 11 -42 „ 9 29. z 81 31. 3 6 18 33. 5.1 64 1 35. — 16 9. a- 6 37. 32 11. 6 * 39 - - 32 41. 20u 2 «- 17 43. -24a;- 2 y- 7 45. --z 16 z n 17. 2 14 2 19.z-io 47.^V 3 21. W - 16 49. l w io w -20 32 23 - 7 " 2 51. 81x- 4 y 28 25. t- 4 53. 32x 30 z- 35 7.1. NEGATIVE EXPONENTS 491 55. -a 16 c' 32 16 X 5 57. y 2 z 3 „9 r 59. ~&? x 3 y s 61. z 5 9 4 U V 63. w 7 -49 65. X 2 67. 56 69. 1 2x 12 71. 3 ~2? 73. 1 81s 36 7c; 1 32y 20 492 CHAPTER 7. RATIONAL EXPRESSIONS 7.2 Scientific Notation We begin this section by examining powers of ten. 10 1 = 10 10 2 = 10-10= 100 10 3 = 10- 10- 10 = 1,000 10 4 = 10- 10- 10- 10= 10,000 Note that the answer for 10 3 is a one followed by three zeros. The answer for 10 4 is a one followed by four zeros. Do you see the pattern? Nonnegative powers of ten. In the expression 10™, the exponent matches the number of zeros in the answer. Hence, 10™ will be a 1 followed by n zeros. EXAMPLE 1. Simplify: 10 9 . Solution. 10 9 should be a 1 followed by 9 zeros. 10 9 = 1,000,000,000 □ You Try It! Simplify: 10 6 Answer: 1,000,000 Next, let's examine negative powers of ten. , 1 10" 1 = — = 0.1 10 ur 10" 10" 100 1 1000 1 10000 0.01 r = 0.001 0.0001 Note that the answer for 10 3 has three decimal places and the answer for 10 -4 contains four decimal places. Negative powers of ten. In the expression 10 _n , the exponent matches the number of decimal places in the answer. Hence, 10 _ra will have n decimal places, the first n — 1 of which are zeros and the digit in the nth decimal place is a 1. 7.2. SCIENTIFIC NOTATION 493 You Try It! EXAMPLE 2. Simplify: 1CT 7 . Simplify: 10~ 5 Solution. 10 should have seven decimal places, the first six of which are zeros, and the digit in the seventh decimal place is a 1. itr 0.0000001 Answer: 0.00001 □ Multiplying Decimal Numbers by Powers of Ten Let's multiply 1.234567 by 10 3 , or equivalently, by 1,000. 1.234567 xlOOO 1234.567000 The sum total of digits to the right of the decimal point in 1.234567 and 1000 is 6. Therefore, we place the decimal point in the product so that there are six digits to the right of the decimal point. However, the trailing zeros may be removed without changing the value of the product. That is, 1.234567 times 1000 is 1234.567. Note that the decimal point in the product is three places further to the right than in the original factor. This observation leads to the following result. Multiplying by a nonnegative power of ten. Multiplying a decimal num- ber by 10™, where n = 0, 1, 2, 3, . . . , will move the decimal point n places to the right. EXAMPLE 3. Simplify: 325.6783 x 10 2 . Solution. Multiplying by 10 2 will move the decimal point two places to the right. Thus: 325.6783 x 10 2 = 32,567.83 You Try It! Simplify: 23.57889 x 10 3 Answer: 23, 578.89 □ EXAMPLE 4. Simplify: 1.25 x 10 5 You Try It! Simplify: 2.35 x 10 4 494 CHAPTER 7. RATIONAL EXPRESSIONS Answer: 23, 500 Solution. Multiplying by 10 5 will move the decimal point two places to the right. In this case, we need to add zeros at the end of the number to accomplish moving the decimal 5 places to the right. 1.25 x 10 5 = 125,000 □ Let's multiply 453.9 by 10 2 , or equivalently, by 0.01. 453.9 xO.01 4.539 The sum total of digits to the right of the decimal point in 453.9 and 0.01 is 3. Therefore, we place the decimal point in the product so that there are 3 digits to the right of the decimal point. That is, 453.9 x 10" 2 = 4.539. Note that the decimal point in the product is two places further to the left than in the original factor. This observation leads to the following result. Multiplying by a negative power of ten. Multiplying a decimal number by 10 _ ™, where n = 1, 2, 3, . . . , will move the decimal point n places to the left. You Try It! Simplify: 3, 854.2 x 10" 1 Answer: 385.42 EXAMPLE 5. Simplify: 14, 567.8 x 10" 3 . Solution. Multiplying by 10 -3 will move the decimal point three places to the left. Thus: 14, 567.8 x 10~ 3 = 14.5678 □ You Try It! Simplify: 2.2 x 10" EXAMPLE 6. Simplify: 4.3 x 10~ 4 . Solution. Multiplying by 10 -4 will move the decimal point four places to the left. In this case, we need to add some leading zeros at the beginning of the number to accomplish moving the decimal 4 places to the left. Answer: 0.022 4.3 x 10" 0.00043 Note also the leading zero before the decimal point. Although .00043 is an equivalent number, the form 0.00043 is preferred in mathematics and science. □ 7.2. SCIENTIFIC NOTATION 495 Scientific Notation Form We start by defining the form of a number called scientific notation. Scientific notation A number having the form a x 10 6 , where b is an integer and 1 < \a\ < 10, is said to be in scientific notation. The requirement 1 < \a\ < 10 says that the magnitude of a must be a least 1 and less than 10. • The number 12.34 x 10 is not in scientific notation because |12.34| = 12.34 is larger than 10. • The number —0.95 x 10 3 is not in scientific notation because | — 0.95| = 0.95 is less than 1. • The number 7.58 x 10 -12 is in scientific notation because |7.58| = 7.58 is greater than or equal to 1 and less than 10. • The number —1.0 x 10 15 is in scientific notation because | — 1.0| = 1.0 is greater than or equal to 1 and less than 10. After contemplating these examples, it follows that a number in scientific no- tation should have exactly one of the digits 1, 2, 3, . . . , 9 before the decimal point. Exactly one, no more, no less. Thus, each of the following numbers is in scientific notation. 4.7 xlO 8 , -3.764 xl0 _1 , 3.2x10°, and - 1.25 x 10 -22 Placing a Number in Scientific Notation To place a number into scientific notation, we need to move the decimal point so that exactly one of the digits 1, 2, 3, . . . , 9 remains to the left of the decimal point, then multiply by the appropriate power of 10 so that the result is equivalent to the original number. You Try It! EXAMPLE 7. Place the number 1,234 in scientific notation. Place the number 54,321 Solution. Move the decimal point three places to the left so that it is posi- tioned just after the 1. To make this new number equal to 1,234, multiply by 10 3 . Thus: 1,234= 1.234 x 10 3 496 CHAPTER 7. RATIONAL EXPRESSIONS Answer: 5.4321 x 10 4 Check: Multiplying by 10 3 moves the decimal three places to the right, so: 1.234 x 10 3 = 1,234 This is the original number, so our scientific notation form is correct. □ You Try It! Place the number 0.0175 EXAMPLE 8. Place the number 0.000025 in scientific notation. Solution. Move the decimal point five places to the right so that it is posi- tioned just after the 2. To make this new number equal to 0.000025, multiply by 10" 5 . Thus: 0.000025 = 2.5 x 10~ 5 Check: Multiplying by 10 -5 moves the decimal five places to the left, so: 2.5 x 10~ 5 = 0.000025 Answer: 1.75 x 10 -2 This is the original number, so our scientific notation form is correct. □ You Try It! Place the number 756.98 x 10" 5 in scientific notation. EXAMPLE 9. Place the number 34.5 x 10 n in scientific notation. Solution. First, move the decimal point one place to the left so that it is positioned just after the three. To make this new form equal to 34.5, multiply by 10 1 . 34.5 x 10~ n = 3.45 x 10 1 x 10~ n Now, repeat the base 10 and add the exponents. 3.45 x 10 -10 Answer: 7.5698 x 10" □ You Try It! Place the number 0.00824 x 10 8 in scientific notation. EXAMPLE 10. Place the number 0.00093 x 10 12 in scientific notation. Solution. First, move the decimal point four places to the right so that it is positioned just after the nine. To make this new form equal to 0.00093, multiply by 10~ 4 . 0.00093 x 10 12 9.3 x 10" 4 x 10 12 7.2. SCIENTIFIC NOTATION 497 Now, repeat the base 10 and add the exponents. = 9.3 x 10 8 Answer: 8.24 x 10 5 Scientific Notation and the Graphing Calculator The TI-84 graphing calculator has a special button for entering numbers in scientific notation. Locate the "comma" key just about the number 7 key on the calculator's keyboard (see Figure 7.1). Just above the "comma" key, printed on the calculator's case is the symbol EE. It's in the same color as the 2nd key, so you'll have to use the 2nd key to access this symbol. □ Figure 7.1: Locate the EE label above the "comma" key on the keyboard. We know that 2.3 x 10 2 = 230. Let's see if the calculator gives the same interpretation. 1. Enter 2.3. 2. Press the 2nd key, then the comma key. This will put E on the calculator view screen. 3. Enter a 2. 4. Press ENTER. The result of these steps is shown in the first image in Figure 7.2. Note that the calculator interprets 2.3E2 as 2.3 x 10 2 and gives the correct answer, 230. You can continue entering numbers in scientific notation (see the middle image in Figure 7.2). However, at some point the numbers become too large and the calculator responds by outputting the numbers in scientific notaiton. You can also force your calculator to display numbers in scientific notation in all situations, by first pressing the MODE key, then selecting SCI on the first 498 CHAPTER 7. RATIONAL EXPRESSIONS 2.3e2 230 2. 3e5 230000 2. 3e10 2.3e10 IriDKHHL HB EDG (i 1 i ]: H E b ? B 5 DEGREE PAR PDL SE9 DDT SIHUL a+bi- rt A $i< HDRIZ G-T inEXTl Figure 7.2: Entering numbers in scientific notation. You Try It! Use the graphing calculator to simplify: (3.42 x 10 6 )(5.86 x 1CT 9 ) line and pressing the ENTER key (see the third image in Figure 7.2). You can return your calculator to "normal" mode by selecting NORMAL and pressing the ENTER key. EXAMPLE 11. Use the graphing calculator to simplify: (2.35 x 1CT 12 )(3.25 x 1(T 4 ) Solution. First, note that we can approximate (2.35 x 10 -12 )(3.25 x 10 -4 ) by taking the product of 2 and 3 and adding the powers of ten. (2.35 x 1(T 12 )(3.25 x 10" 4 ) w (2 X 1(T 12 )(3 x 1(T 4 ) Approximate: 2.35 « 2 and 3.25 « 3. w 6 x 10~ 16 2 • 3 = 6 and 10" 12 ■ 10" 4 = 10" 16 . The graphing calculator will provide an accurate answer. Enter 2.35E-12, press the "times" button, then enter 3.25E-4 and press the ENTER button. Be sure to use the "negate" button and not the "subtract" button to produce the minus sign. The result is shown in Figure 7.3. 2.35e-12*3.25e-4 7.6375E-16 Figure 7.3: Computing (2.35 x 1(T 12 )(3.25 x 10" Answer: 2.00412 x 10" Thus, (2.35 x 10~ 12 )(3.25 x 10~ 4 ) close to our estimate of 6 x 10~ 16 . 7.6375 x 10" lb . Note that this is fairly □ 7.2. SCIENTIFIC NOTATION 499 Reporting your answer on your homework. After computing the answer to Example 11 on your calculator, write the following on your homework: (2.35 x 1CT 12 )(3.25 x 10" 4 ) = 7.6375 x 1CT 16 Do not write 7.6375E-16. You Try It! EXAMPLE 12. Use the graphing calculator to simplify: 6.1 x 1(T 3 (2.7 x 10 4 )(1.1 x 10 8 ) Use the graphing calculator to simplify: 2.6 x 10 4 (7.1 x 10" 2 )(6.3x 10 7 ) Solution. Again, it is not difficult to produce an approximate answer. 6.1 x 10~ 3 (2.7 x 10 4 )(1.1 x 10 8 ) 6 x 10" 3 ~ (3 x 10 4 )(1 x 10 8 ) Jx 10" 3 " 3 x 10 12 - 6 10~ 3 ~ 3 'TniT .1«6, 2.7 « 3, and 1.1 w 1. 10 2 x 10 -15 3 • 1 = 3 and 10 4 • 10 8 = 10 ac a c bd = b'd' 6 10~ 3 - = 2 and — — = 10" 3 10 12 12 -15 Let's get a precise answer with our calculator. Enter the numerator as 6.1E- 3, then press the "division" button. Remember that we must surround the denominator with parentheses. So press the open parentheses key, then enter 2.7E4. Press the "times" key, then enter 1.1E8. Press the close parentheses key and press the ENTER button. The result is shown in Figure 7.4. 6.1e-3/<2.7e4*1. 1eS> 2.053S7205E-15 Figure 7.4: Computing 6.1 x 10" 3 /(2.7 x 10 4 x 1.1 x 10 Thus, 6.1 x 10~ 3 /(2.7 x 10 4 x 1.1 x 10 8 ) = 2.05387205 x 10" 15 . Note that this is fairly close to our estimate of 2 x 10 -15 . Answer: 5.8126537 x 10" □ 500 CHAPTER 7. RATIONAL EXPRESSIONS You Try It! The mass of the International Space Station is 450,000 kg, and its average distance to the center of the earth is 387,000 m. Find the force of attraction between the earth and the station (in newtons (N)). EXAMPLE 13. Isaac Newton's universal law of gravitation is defined by the formula GmM where F is the force of attraction between two objects having mass m and M, r is the distance between the two objects, and G is Newton's gravitational constant defined by: G = 6.67428 x 10" 11 N(m/kg) 2 Given that the mass of the moon is 7.3477 x 10 22 kilograms (kg), the mass of the earth is 5.9736 x 10 24 kilograms (kg), and the average distance between the moon and the earth is 3.84403 x 10 8 meters (m), find the force of attraction between the earth and the moon (in newtons (N)). Solution. Plug the given numbers into Newton's universal law of gravitation. GmM F = - — F (6.673 x 10~ n )(7.3477 x 10 22 )(5.9736 x 10 24 ) (3.84403 x 10 8 ) 2 Enter the expression into your calculator (see Figure 7.5) as: (6.673E-11*7.3477E22*5.9736E24)/(3.84403E8)A2. Answer: 1.20 x 10 9 N C6.673E -11*7. 347 7e22*5.9736e24V (3.84403 £8^2 1.9821 43728 e2S Figure 7.5: Computing force of attraction between earth and the moon. Hence, the force of attraction between the earth and the moon is approxi- mately 1.98 x 10 20 newtons (N). □ You Try It! The star Sirius is 8.58 light-years from the earth. How many miles from the earth is Sirius? EXAMPLE 14. The closest star to the earth is Alpha Centauri, 4.37 light- years from the earth. A light-year is the distance that light will travel in 7.2. SCIENTIFIC NOTATION 501 one-year's time. The speed of light is 186,000 miles per second. How many miles from the earth is Alpha Centauri? Solution: Because the speed of light is measured in miles per second, let's first compute the number of seconds in 4.37 years. Because there are 365 days in a year, 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute, we can write: day hr min s 4.37 yr = 4.37yr x 365 — x 24 - — x 60 - — x 60 — — yr day hr min day' faf rain s = 4.37 yr x 365 — - x 24 -— x 60 ±=-r x 60 — ^ yf pay .far jam Note how the units cancel, indicating that the final answer is in seconds. With our calculator mode set to scientific notation (see the image on the right in Figure 7.2), we multiply the numbers to get the result shown in Figure 7.6. Rounding, the number of seconds in 4.37 years is approximately 1.38 x 10 8 seconds. Next, we compute the distance the light travels in 4.37 years. Using the fact that the distance traveled equals the speed times the time traveled, we have: Distance = Speed x Time = 1.86 x 10 5 — • 1.38 x 10 8 s s = 1.86 x 10 5 ^ • 1.38 x lO 8 * 1 r Note how the units cancel, indicating that our answer is in miles. Again, with our calculator set in scientific notation mode, we compute the product of 1.86 x 10 5 and 1.38 x 10 8 . The result is shown in the image on the right in Figure 7.6. 4.37*365*24*60*6 1. 3781232 e8 1.86e5*1.33e8 2.5668e13 Figure 7.6: Computing the distance to Alpha Centauri in miles. Thus, the star Alpha Centauri is approximately 2.5668 x 10 13 miles from the earth, or 2.5668 x 10 13 miles w 25, 668, 000, 000, 000 miles, pronounced "twenty-five quadrillion, six hundred sixty-eight trillion miles." Answer: « 5.2425 x 10 13 miles □ 502 CHAPTER 7. RATIONAL EXPRESSIONS **> t* t* Exercises ■** ■*$ •*$ In Exercises 1-8, write each of the following in decimal format. 1. 10" 4 5. 10 8 2. 1(T 13 6. 10 14 3. 1CT 8 7. 10 7 4. 10" 9 8. 10 9 In Exercises 9-16, write each of the following in decimal format. 9. 6506399.9 x 10~ 4 13. 440906.28 x 10" 4 10. 19548.4 x 10~ 2 14. 9147437.4 x 10~ 4 11. 3959.430928 x 10 2 15. 849.855115 x 10 4 12. 976.841866 x 10 4 16. 492.4414 x 10 3 In Exercises 17-24, convert each of the given numbers into scientific notation. 17. 390000 21. 0.81 18. 0.0004902 22. 83400 19. 0.202 23. 0.0007264 20. 3231 24. 0.00395 In Exercises 25-32, convert each of the given expressions into scientific notation. 25. 0.04264 x 10~ 4 29. 30.04 x 10 5 26. 0.0019 x 10" 1 30. 76000 x 10" 1 27. 130000 x 10 3 31. 0.011 x 10 1 28. 738 x lO" 1 32. 496000 x 10~ 3 7.2. SCIENTIFIC NOTATION 503 In Exercises 33-38, each of the following numbers are examples of numbers reported on the graphing calculator in scientific notation. Express each in plain decimal notation. 33. 1.134E-1 34. 1.370E-4 35. 1.556E-2 36. 1.802E4 37. 1.748E-4 38. 1.402E0 In Exercises 39-42, first, use the technique of Example 11 to approximate the given product without the use of a calculator. Next, use the MODE button to set you calculator in SCI and FLOAT mode, then enter the given product using scientific notation. When reporting your answer, report all digits shown in your calculator view screen. 39. (2.5 x 10 _1 )(1.6 x 1(T 7 ) 40. (2.91 x 10" 1 )(2.81 x 1(T 4 ) 41. (1.4 x 10 7 )(1.8 x 1CT 4 ) 42. (7.48 x 10 7 )(1.19 x 10 6 ) In Exercises 43-46, first, use the technique of Example 12 to approximate the given quotient without the use of a calculator. Next, push the MODE button, then highlight SCI mode and press ENTER. Move your cursor to the same row containing the FLOAT command, then highlight the number 2 and press ENTER. This will round your answers to two decimal places. Press 2nd MODE to quit the MODE menu. With these settings, enter the given expression using scientific notation. When entering your answer, report all digits shown in the viewing window. 3.2 x 1CT 5 43. = 2.5 x 10- 7 6.47 x 1CT 5 44 1.79 x 10 8 45 46 5.9 x 10 3 2.3 x 10 5 8.81 x 10" 9 3.06 x lO" 1 47. Overall the combined weight of biolog- ical material - animals, plants, insects, crops, bacteria, and so on - has been es- timated to be at about 75 billion tons or 6.8 x 10 13 kg (http: //en.wikipedia. org/wiki/Nature). If the Earth has mass of 5.9736 x 10 24 kg, what is the percent of the Earth's mass that is made up of biomass? 48. The Guinness World Record for the longest handmade noodle was set on March 20, 2011. The 1,704-meter- long stretch of noodle was displayed during a noodle-making activity at a square in Southwest China's Yunnan province. Meigan estimates that the average width of the noodle (it's diameter) to be the same as her index finger or 1.5 cm. Using the volume formula for a cylinder (V = irr 2 h) estimate the volume of the noodle in cubic centimeters. 504 CHAPTER 7. RATIONAL EXPRESSIONS 49. Assume there are 1.43 x 10 6 miles of paved road in the United States. If you could travel at an average of 65 miles per hour nonstop, how many days would it take you to travel over all of the paved roads in the USA? How many years? 50. The population of the USA in mid-2011 was estimated to be 3.12 x 10 8 people and the world population at that time to be about 7.012 x 10 9 people. What percent- age of the world population live in the USA? j*- j*- j*- Answers •*$-*$■*$ 1. 0.0001 3. 0.00000001 5. 100,000,000 7. 10,000,000 9. 650.63999 11. 395943.0928 13. 44.090628 15. 8498551.15 17. 3.9 x 10 5 19. 2.02 x lO" 1 21. 8.1 x lO" 1 23. 7.264 x 10~ 4 25. 4.264 x 10~ 6 27. 1.3 x 10 8 29. 3.004 x 10 6 31. 1.1 x 10" 1 33. 0.1134 35. 0.01556 37. 0.0001748 39. 4 x 10~ 8 41. 2.52 x 10 3 43. 1.28 x 10 2 45. 2.57 x 10~ 2 47. 1.14 x 10" 11 49. 916.7 days, 2.5 yr 7.3. SIMPLIFYING RATIONAL EXPRESSIONS 505 7.3 Simplifying Rational Expressions Any time you divide a polynomial by a second polynomial, you form what is known as a rational expression. Rational expression. The expression P{x) q(x) } where p(x) and q(x) are polynomials, is called a rational expression. Readers are strongly encouraged to review the material on fractions presented in Section 3 of Chapter 1. You will find that material quite helpful for this section. For example, each of the following is a rational expression. ■2 , s x + 3 , 2x c) v 3.7.' b) 2a; -4 In example a) , the rational expression is composed of a binomial over a mono- mial. Example b) is constructed by dividing a binomial by a trinomial. Ex- ample c) is composed of a monomial over a monomial, the type of rational expression that will gain the most attention in this section. Multiplying and Dividing Rational Expressions We will concentrate on rational expressions with monomial numerators and denominators. Recall that to form the product of two rational numbers, we simply multiply numerators and denominators. The same technique is used to multiply any two rational expressions. Multiplying rational expressions. Given a/b and c/d, their product is defined as: a c ac ~b"d = bd Remember, you need only multiply numerators and denominators. For exam- ple: x 2 3'y 2.7' 3? 2a 5a 36 1 ' g^3 10a 2 27&5' and x 2y 3x 4y2 3x 2 8^ Of course, as the next example shows, sometimes you also need to reduce your answer to lowest terms. EXAMPLE 1. Simplify: You Try It! Simplify: 9 x x 2 6 506 CHAPTER 7. RATIONAL EXPRESSIONS Answer: — 2x Solution: Multiply numerators and denominators. 2 x 2 _ 2x b x 4 4x 3 Now, there several different ways you can reduce this answer to lowest terms, two of which are shown below. You can factor numerator and de- Or you can write the answer as a nominator, then cancel common product, repeat the base and sub- factors. tract exponents. 2a; 5 4a; 3 2 ■ x ■ x ■ x ■ x ■ x 2 ■ 2 ■ x ■ x ■ x % ■ ft ■ ft ■ ft ■ X ■ X _x^ 2 As dividing by 2 is the same as multiplying by 1/2, these answers are equivalent. Also, note that the right-hand method is more efficient. □ 2a; 5 2 X 5 4x 3 "" 4 X 3 = 1 2 x 5 - = 1 2 r Recall that when dividing fractions, we invert the second fraction and mul- tiply. Dividing rational expressions. Given a/b and c/d, their quotient is defined as: a b c ~ d ~ a d b'~c ad ~b~c You Try It! Simplify: 3y ^ r a; 3 4a; 2 4 EXAMPLE 2. Simplify: — -=- At V 2y^ Solution: Invert, then multiply. 1 A X _ X x 2 2y 2 y ' 2y 2 y x A 2x 2 y 2 x 4 y Now, there several different ways you can reduce this answer to lowest terms, two of which are shown below. 7.3. SIMPLIFYING RATIONAL EXPRESSIONS 507 You can factor numerator and de- nominator, then cancel common factors. 2x 2 y 2 x A y Or you can write the answer as a product, repeat the base and sub- tract exponents. 2 ■ x ■ x ■ y !J x ■ x ■ X ■ X y y 2„,2 1x z y x 4 y x 4 y 1 ft ■ ft ■ X ■ X ■ 2y 2x- 2 y l 2y In the last step, x~ 2 is the same as 1/x 2 , then we multiply numerators and denominators. Note that the right-hand method is more efficient. Answer: 12 x 2 y Adding and Subtracting Rational Expressions First, recall the rules for adding or subtracting fractions that have a "common" denominator. Adding rational expressions. Given a/c and b/c, their sum is defined as: a b a + b c c c That is, add the numerators and place the result over the common denominator. The following examples each share a common denominator. We add the nu- merators, then place the result over the common denominator. □ 5 1 _ 6 2 3_5 7 7 7' x x x' and x 3y x + 3y y y y 3a: 2v EXAMPLE 3. Simplify: — + — xy xy Solution: Add the numerators, placing the result over the common denomi- nator. 3a; 2y 3a; + 2y xy xy xy You Try It! Simplify: — 1 r— x^y x z y Answer: 4a; + by 2 x 2 y □ 508 CHAPTER 7. RATIONAL EXPRESSIONS You Try It! Simplify: 8.r 2x 3yz 2 3yz 2 Subtracting rational expressions. Given a/c and b/c, their difference is defined as: a b a — b c c c That is, subtract the numerators and place the result over the common denom- inator. The following examples each share a common denominator. We subtract the numerators, then place the result over the common denominator. 5a T 3o T 2a and 3;r xy xy 3x — by xy As the next example shows, sometimes you may have to reduce your answer to lowest terms. EXAMPLE 4. Simplify: — - — Solution: Subtract the numerators, placing the result over the common denominator. hxy 3xy bxy — 3xy ~z 2z ~ 2~z _ 2xy 2z To reduce to lowest terms, divide both numerator and denominator by 2. Answer: 2x yz 2 xy_ z □ The Least Common Denominator When adding or subtracting, if the rational expressions do not share a com- mon denominator, you must first make equivalent fractions with a common denominator. Least common denominator. If the fractions a/b and c/d do not share a common denominator, then the least common denominator for b and d is defined as the smallest number (or expression) divisible by both b and d. In symbols, LCD(6, d) represents the least common denominator of b and d. 7.3. SIMPLIFYING RATIONAL EXPRESSIONS 509 You Try It! EXAMPLE 5. Simplify: 1 1 J 6 9 Simplify: — 5:)' Solution: The smallest number divisible by both 6 and 9 is 18; i.e., LCD(6, 9) = 18. We must first make equivalent fractions with a common denominator of 18. x 2x 6 + Y a; 3 2a; ~ 6 ' 3 + ~9~ 3a; 4a; ~ 18 + 18 2 2 We can now add the numerators and put the result over the common denomi- nator. - 7 JL ~ 18 Answer: 29.x ~2A □ You Try It! EXAMPLE 6. Simplify: JL - JL Solution: The smallest expression divisible by both 8a; and 12a; is 24a;; i.e., LCD(8a;, 12a;) = 24a;. We must first make equivalent fractions with a common denominator of 24a;, then place the difference of the numerators over the common denominator. Simplify: 8y lOy y 8x ' y _ y 3 y " 12a; ~ 8a; ' 3 12a; _ 3y 2y 24a; 24a; y 24a; 2 2 Answer: 40y D In Example 5, it was not difficult to imagine the smallest number divisible by both 6 and 9. A similar statement might apply to Example 6. This is not the case in all situations. EXAMPLE 7. Simplify: 21L 72 y 108 You Try It! Simplify: 7x 36 3.r 40 510 CHAPTER 7. RATIONAL EXPRESSIONS Solution: In this example, it is not easy to conjure up the smallest number divisible by both 72 and 108. As we shall see, prime factorization will come to the rescue. 72 108 27 Thus, 72 = 2 3 ■ 3 2 and 108 = 2 2 • 3 3 . Procedure for finding the least common denominator (LCD). To find the least common denominator for two or more fractions, proceed as follows: 1. Prime factor each denominator, putting your answers in exponential form. 2. To determine the LCD, write down each factor that appears in your prime factorizations to the highest power that it appears. Following the procedure above, we list the prime factorization of each denom- inator in exponential form. The highest power of 2 that appears is 2 3 . The highest power of 3 that appears is 3 3 . Answer: 43a; 360 72 108 LCD 2 3 -3 2 2 2 -3 3 2 3 -3 3 Therefore, the LCD is 2 3 • 3 3 hy 72 " y _5y 3 y ' 108 ~~ 72 ' 3 108 _ 15y 2y 216 216 _ 13y 216 2 ' 2 Prime factor 72. Prime factor 108. Highest power of 2 is 2 . Highest power of 3 is 3 . 27 or 216. Hence: Make equivalent fractions. Simplify. Subtract numerators. □ 7.3. SIMPLIFYING RATIONAL EXPRESSIONS 511 EXAMPLE 8. Simplify: 7 11 15xy 2 20a; 2 Solution: Prime factor each denominator, placing the results in exponential form. 15xy = 3 • 5 ■ x ■ y 20a; 2 = 2 2 ■ 5 • x 2 To find the LCD, list each factor that appears to the highest power that it appears. You Try It! Simplify: 11 7x I8x 2 y 30xy 2 „,2 LCD = 2 Z ■ 3 • 5 • x z ■ y Simplify. LCD = 60a;V After making equivalent fractions, place the difference of the numerators over this common denominator. 7 11 7 Ax 11 3y 2 I5xy 2 20x 2 lbxy 2 Ax 20x 2 3y 2 28a; 33y 2 60x 2 y 2 60x 2 y 2 28a; - 33y 2 60x 2 y 2 Answer: 55 + 21a; J 90x 2 y D Dividing a Polynomial by a Monomial We know that multiplication is distributive with respect to addition; that is, a(b + c) = ab + ac. We use this property to perform multiplications such as: x 2 (2a; 2 -3a;- 2a; 4 - 3x d 8x z However, it is also true that division is distributive with respect to addition. 512 CHAPTER 7. RATIONAL EXPRESSIONS You Try It! Divide 9x 3 - by 3a; 2 . 8a; — 6a; For example, note that 4 + 6 _ 4 6 2 ~ 2 + 2' This form of the distributive property can be used to divide a polynomial by a monomial. EXAMPLE 9. Divide x 2 - 2x - 3 by x 2 . Solution: We use the distributive property, dividing each term by x 2 . Answer: 3a; H 3 x Now we reduce each term of the last result to lowest terms, canceling common factors. 2 3 □ You Try It! Divide -Ax 2 + 6x - 9 by 2a; 4 . 2 3 9 Answer: -\ EXAMPLE 10. Divide 2x 3 - 3a; + 12 by 6a; 3 . Solution: We use the distributive property, dividing each term by 6a; 3 . 2a; 3 - 3a; + 12 _ 2x 3 3x 12 6a; 3 6a; 3 6x 3 6a; 3 Now we reduce each term of the last result to lowest terms, canceling common factors. 1 1 2^2 □ 7.3. SIMPLIFYING RATIONAL EXPRESSIONS 513 **>**■**> Exercises ■** * «•* In Exercises 1-8, simplify each of the given experssions. 12 s 5 1. T 2 ' 9 2. 6 T 4 a: 2 ' 10 12 v 4 3. y 3 10 4. 10 ¥' i 5 12 s 5 9s 2 b. t 4 ' " t 2 s 2 6s 4 6. £ 2 ' ' t 4 6 4 96 2 7. c 4 ' "T 2 " 6 5 86 2 8. c 4 ' ' c 2 In Exercises 9-14, simplify each of the given expressions. 9. 10s 19s ~T8~ + Ts" 10. Uy lOy 11. 5 17 9c ~ 9c 12. 19 147 ~ 17 147 13. 8a; \byz 16a; 15j/z 1/1 17a 9a 206c 206c In Exercises 15-20, simplify each of the given expressions. 15. 9z 5z 10 + T 16. 7u llu 17. 3 4 10w 5v 9 18. 7 lOv 2v 8r 19. 5st 9r lOst 7a; 3.r 6y^ 2 W z In Exercises 21-32, simplify each of the given expressions. 11 5 5 17 21. k + —^ 23. 18rs 2 24r 2 s ' 24rs 2 36r 2 s 5 13 13 19 22. 24. Yluw 1 hAu 2 w 54ww 2 2Av 2 w 514 CHAPTER 7. RATIONAL EXPRESSIONS 7 11 25. - H - 29. 36y 3 48z 3 ™ 19 5 26 36^ + 48? 3 °- 27.^ + ^- 31. 48w 3 36u> 3 7 17 28 1 32 72r 3 48s 3 50rs 40st In Exercises 33-48, use the distributive property to divide each term in the numerator by the term in the denominator. 11 9 50xy AOyz 9 13 50rs AQst 19 17 50a6 406c 9 11 QQ 6w + 12 3x 2_ 8x _ 9 oo. 41. 34. —^— 42. 25w + 45 35. 43. 5 16a; + 4 36. 44. „„ 2s " 4 37. 45. s „„ 7r ~ 8 38. 46. r 3r-5 39. 47. 40. 48. u lOur 1. 4s 3 IT 3. 6c t 5. s 3 9£ 2 X 2 46 2 - -56- 8 6 2 2a; 2 — 3a; - -6 a; 2 6u 2 — 5w - -2 U 2 12t 2 + 2t- -16 I2t 2 186 2 : + 96 - 15 186 2 4s 2 + 2s- 10 4s 2 Ww 2 + 12w - 2 5*- $*> £»■ Answers ■*? -as -as 7.il 9c 2 -I 4 11. 3c 7.3. SIMPLIFYING RATIONAL EXPRESSIONS 515 Sx 13. 5yz 17 z 15. 5 17.-1 2v 5r 19. 2st 44r + 15s 21 " 72r 2 s 2 15r + 34s 23. 72r 2 s 2 3 i qq„,3 25 28^ + 33y 27. 144y3 z 3 15w 3 + 52v 3 144v 3 uj 3 44z - 45a; 29. 200xyz 31. 76c :-85a 200abc 33. 2v- f 4 35. 5m + 9 37. 2- 4 s 39. 3- 5 r 41. 3- 8 9 X X 2 43. 2- 3 6 X X 2 45. 1 + 1 4 6i ~ 3T 2 47. 1 + 1 5 Ys~ 2^ 516 CHAPTER 7. RATIONAL EXPRESSIONS 7.4 Solving Rational Equations In Section 3 of Chapter 2, we showed that the most efficient way to solve an equation containing fractions was to first clear the fractions by multiplying both sides of the equation by the least common denominator. For example, given the equation 1 1 _ 1 2 X+ 3 = 4' we would first clear the fractions by multiplying both sides by 12. 12 ri i pi ~x + - = — [2 3 L 4 J - 12 Qx + 4 = 3 This procedure works equally well when the the denominators contain a vari- able. You Try It! Solve for x: 1 6 EXAMPLE 1. Solve for a;: 1 - - = — Solution: The common denominator is x 2 . We begin by clearing fractions, multiplying both sides of the equation by x 2 . 1-1 = 1 x x 2 ?,' r 3 1 1 - _ — X _ X 2 Original equation. Multiply both sides by x Now we use the distributive property. x 2 [1] - x 2 Distribute x . Now we cancel common factors and simplify. 2x = 3 Cancel. Simplify. The resulting equation is nonlinear (x is raised to a power larger than 1 . Make one side zero, then factor. x z - 2x - 3 = (a:-3)(a; + l) = Nonlinear. Make one side zero. Factor. 7.4. SOLVING RATIONAL EQUATIONS 517 Use the zero product property to complete the solution. Either the first factor is zero or the second factor is zero. a; — 3 = or £ + 1 = x = 3 x = — 1 Hence, the solutions are x = —1 and x = 3. Check. Substitute —1 for x, then 3 for x in the original equation and simplify. 1-2 * x x 2 i-2 = A X X 2 1 (-1) (-1) 2 1 + 2 = 3 3 = 3 1 1 (3) (S) 2 2 _ 3 ~ 3 ~ 9 1 1 3 ~ 3 Note that both result in true statements, showing that both x = — 1 and x = 3 check in the original equation. Answer: 2, 4 □ 22 29 EXAMPLE 2. Solve for x: 6- — = — x 2 x Solution: The common denominator is x 2 . 22 _ 29 72 6 x- :r x x 2 [6] 22" [291 6 TT =: X 2 . x . [22" [291 r." ::i_: X 2 X X 6x 2 - 22 = 29x Original equation. Multiply both sides by x . Distribute x . Cancel and simplify. You Try It! 7 Solve for x: — This last equation is nonlinear. Make one side zero. 6x 2 - 29x - 22 = The integer pair 4 and —33 has product ac = —132 and sum b = —29. Break up the middle term into a sum using this pair, then factor by grouping. 6x 2 + Ax - 33x - 22 = 2x(3x + 2)- ll(3x + 2) = (2x- ll)(3x + 2) = 30 x 518 CHAPTER 7. RATIONAL EXPRESSIONS Finally, use the zero product property to write: 2x 11 = or 3a; + 2 = 2a; = 11 3a; = -2 11 X = Y 2 Answer: -1/4, -7/2 Check: Let's check these solutions with our calculators. Enter 11/2, push the STO^ button, push the X,T,6>,n button and the ENTER key (see the calculator screen on the left in Figure 7.7). Next, enter the left-hand side of the equation as 6-22/X~2 and press ENTER. Enter the right-hand side of the equation as 29/X and press ENTER. The results are the same (see the calculator screen on the left in Figure 7.7). This verifies that 11/2 is a solution of 6— 22/a; 2 = 29/a;. The calculator screen on the right in Figure 7.7 shows a similar check of the solution x = —2/3. ll/2-»X 5.5 6-22/X A 2 5.272727273 29/X 5.272727273 -2/3-»X -.6666666667 6-22/X A 2 -43.5 29/X -43.5 22 29 Figure 7.7: Checking the solutions of 6 ~ = — ■ Solving Rational Equations with the Graphing Calculator □ Let's use the graphing calculator to solve an equation containing rational ex- pressions. You Try It! Solve the equation 2 5 x 12 both algebraically and graphically, then compare your solutions. EXAMPLE 3. Consider the following equation: x x Solve the equation algebraically, then solve the equation graphically using your graphing calculator. Compare your solutions. 7.4. SOLVING RATIONAL EQUATIONS 519 Algebraic solution: First, an algebraic approach. Multiply both sides of the equation by the common denominator x 2 . 2-£ = A x x A x 2 \2\ 9" ' 5 " 2 — x _ X 1 2 "9" ' 5 " X — ^ X _ X 2 2x z - 9x Original equation. Multiply both sides by x 2 Distribute x . Cancel and simplify. The last equation is nonlinear. Make one side zero. 2x z - 9a; Make one side zero. The integer pair —10 and 1 have product equaling ac = —10 and sum equaling b = —9. Break up the middle term using this pair, then factor by grouping. 2x 2 - 10a; + x - 5 = 2a;(x-5) + l(»-5) =0 (2x+l)(cc-5) =0 — 10a; + x = —9x. Factor by grouping. Factor out x — 5. Now use the zero product property to write: 2a; + 1 = or x - -5 = 2a; = -1 x = 5 1 X = ~2 re x = —1/2 and x = 5. Hence, the solutions are x Graphical solution: We could load each side of the equation separately, then use the intersect utility to find where the graphs intersect. However, in this case, it's a bit easier to make one side of the equation zero, draw a single graph, then note where the graph crosses the a;-axis. 9 5 2-- = — x x A 9 5 2 - = x x z Original equation. Make one side zero. Load the left-hand side of the equation into Yl as 2-9/X-5/XA2 (see the image on the left in Figure 7.8), then select 6:ZStandard from the ZOOM menu to produce the image at the right in Figure 7.8. Next, the solutions of 9 5 2 x x z 520 CHAPTER 7. RATIONAL EXPRESSIONS Ploti Plots Plots \ViB2-9/X-5^T2 ■xVs = ■xVfi = \r Figure 7.8: Sketch the graph of y = 2 9 x are found by noting where the graph of y = 2 — 9/a: — 5/x 2 cross the a;-axis. Select 2 : zero from the CALC menu. Use the arrow keys to move the cursor to the left of the first x-intercept, then press ENTER to set the "Left bound." Next, move the cursor to the right of the first x-intercept, then press ENTER to set the "Right bound." Finally, leave the cursor where it is and press ENTER to set your "Guess." The calculator responds with the result shown in the figure on the left in Figure 7.9. Repeat the zero-finding procedure to capture the coordinates of the second x-intercept (see the image on the right in Figure 7.9). \ 2*K0 K=".£ V 2*K0 V=(i Figure 7.9: Locating the zeros of y = 2 Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with x and y, respectively (see Figure 7.10). • Place your WINDOW parameters at the end of each axis (see Figure 7.10). • Label the graph with its equation (see Figure 7.10). • Drop dashed vertical lines through each x-intercept. Shade and label the cc-values of the points where the dashed vertical line crosses the x- axis. These are the solutions of the equation 2 — 9/x — 5/x 2 = (see Figure 7.10). 7.4. SOLVING RATIONAL EQUATIONS 521 y = 2 - 9/x - 5/x 2 Figure 7.10: Reporting your graphical solution on your homework. Thus, the calculator is reporting that the solutions of 2 — 9/z — 5/x 2 = are x = —0.5 and x = 5, which match the algebraic solutions x = —1/2 and x = 5. Numerical Applications Let's apply what we've learned to an application. Answer: —4, 3/2 : f K=-4 1 v=o : t /~ K=1.E I □ You Try It! EXAMPLE 4. The sum of a number and its reciprocal is 41/20. Find the The sum of a number and its number. reciprocal is 53/14. Find the number . Solution: In the solution, we address each step of the Requirements for Word Problem Solutions. 1. Set up a variable dictionary. Let x represent the unknown number. 2. Set up an equation. If the unknown number is x, then its reciprocal is 1/x. Thus, the "sum of a number and its reciprocal is 41/20" becomes: x + 1 _ 41 x ~ 20 522 CHAPTER 7. RATIONAL EXPRESSIONS 3. Solve the equation. Clear the fractions by multiplying both sides by 20a;, the least common denominator. 1 41 x H = — Model equation. x 20 11 [41 20a; x + - = — 20a; Multiply both sides by 20a;. 1" [411 x+ - = X _ L20J \Y [41 "1 211.7: — = X _ L20J 20a; \x] + 20a; - = — 20a; Distribute 20a;. 20x + 20 = 41a; Cancel and simplify. The equation is nonlinear. Make one side zero. 20x 2 - 41a; + 20 = Make one side zero. The integer pair —16 and —25 has product ac = 400 and sum 6 = —41. Break up the middle term in the last equation into a sum of like terms using this pair, then factor by grouping. 20a; 2 - 16a; - 25a; + 20 = - 16a; - 25a; = -41a;. 4a;(5a; — 4) — 5(5a; — 4) = Factor by grouping. (4a; - 5) (5a; - 4) = Factor out 5a; - 4. We can now use the zero product property to write: 4a; - 5 = or 5a; -4 = 4a; = 5 5a; = 4 5 X= 4 4 X= 5 Answer: 2/7, 7/2 4. Answer the question. There are two possible numbers, 5/4 and 4/5. 5. Look back. The sum of the unknown number and its reciprocal is supposed to equal 41/20. The answer 5/4 has reciprocal 4/5. Their sum is: 5 4 _ 16 25 4 + 5 ~~ 20 + 20 41 ~ 20 Thus, 5/4 is a valid solution. The second answer 4/5 has reciprocal 5/4, so it is clear that their sum is also 41/20. Hence, 4/5 is also a valid solution. □ 7.4. SOLVING RATIONAL EQUATIONS 523 **>**■**> Exercises ■** * «•* In Exercises 1-8, solve the equation. 1. 26 x = 11 + — a; 2. „ 60 x = 7H 3. 12 27 x a; 2 4. 1 6 7 x x z 10 11 5. 1- X X 2 20 96 6. 1- X X 2 44 7. x = = 7+ — a; 99 8. a; = = 2 + — a; In Exercises 9-16, solve the equation. 9. 12a; = 97 a; 10. 7x = -19 a: 11. 19 3 20+ — = - x x z 12. 8 1 33 - - = -s x x z 13. 11 8x = 19 X 14. 3 28x = 25 X 15. 40 + ^ = 1 x x z 16. 11 1 18 + — = ? x x z In Exercises 17-20, solve each equation algebraically, then use the calculator to check your solutions. 17. 36x=-13 19. 14a; = 9 x x 10 20 18. 9x = 43+— 20. 3x = 16 x x In Exercises 21-24, solve the equation algebraically, then solve the equation using the graphing calcu- lator using the technique shown in Example 3. Report your solution using the Calculator Submission Guidelines demonstrated in Example 3. I 12 44 21.1-- = — 23. 2x = 3 + — X X A X II 28 4 22.1 + — = 24. 2x = 9-- xx 2 - X 524 CHAPTER 7. RATIONAL EXPRESSIONS 25. The sum of a number and its reciprocal is 5/2. Find the number. 26. The sum of a number and its reciprocal is 65/8. Find the number. 27. The sum of a number and 8 times its recip- rocal is 17/3. Find all possible solutions. 28. The sum of a number and 4 times its recip- rocal is 17/2. Find all possible solutions. j*- j*- ?*- Answers •** ■*$ ■** 1. -2, 13 3. 3, 9 5. 11, -1 7. -4, 11 9. 8, 1/12 11. -3/4, -1/5 13. 11/8, 1 15. -1/4, 1/10 17. -1/9, -1/4 19. 1/2, 1/7 21. -3, 4 23. -4, 11/2 25. 2, 1/2 27. 3, 8/3 7.5. DIRECT AND INVERSE VARIATION 525 7.5 Direct and Inverse Variation We start with the definition of the phrase "is proportional to." Proportional. We say that y is proportional to x if and only if y = kx, where k is a constant called the constant of proportionality. The phrase "y varies directly as x" is an equivalent way of saying "y is proportional to x." Here are a few examples that translate the phrase "is proportional to." • Given that d is proportional to t, we write d = kt, where k is a constant. • Given that y is proportional to the cube of x, we write y = kx 3 , where k is a constant. • Given that s is proportional to the square of t, we write s = kt 2 , where k is a constant. We are not restricted to always using the letter k for our constant of propor- tionality. You Try It! EXAMPLE 1. Given that y is proportional to x and the fact that y = 12 Given that y is proportional when x = 5, determine the constant of proportionality, then determine the to x and that y = 21 when value of y when x = 10. x = 9, determine the value of Solution: Given the fact the y is proportional to x, we know immediately that y = kx, where k is the proportionality constant. Because we are given that y = 12 when x = 5, we can substitute 12 for y and 5 for x to determine k. y = kx y is proportional to x. 12 = fc(5) Substitute 12 for y, 5 for x. 12 — = k Divide both sides by 5. 5 Next, substitute the constant of proportionality 12/5 for k in y = kx, then substitute 10 for x to determine y when x = 10. 12 y = — x Substitute 12/5 for k. 5 12 y = — (10) Substitute 10 for x. 5 y = 24 Cancel and simplify. Answer: 63 □ 526 CHAPTER 7. RATIONAL EXPRESSIONS You Try It! A ball is dropped from the edge of a cliff on a certain planet. The distance s the ball falls is proportional to the square of the time t that has passed since the ball's release. If the ball falls 50 feet during the first 5 seconds, how far does the ball fall in 8 seconds? EXAMPLE 2. A ball is dropped from a balloon floating above the surface of the earth. The distance s the ball falls is proportional to the square of the time t that has passed since the ball's release. If the ball falls 144 feet during the first 3 seconds, how far does the ball fall in 9 seconds? Solution: Given the fact the s is proportional to the square of t, we know immediately that s = kt 2 , where k is the proportionality constant. Because we are given that the ball falls 144 feet during the first 3 seconds, we can substitute 144 for s and 3 for t to determine the constant of proportionality. s = kt 2 144 = fc(3) 2 144 = 9fc 16 = k s is proportional to the square of t. Substitute 144 for s, 3 for t. Simplify: 3 2 = 9. Divide both sides by 9. Next, substitute the constant of proportionality 16 for k in s = kt 2 , and then substitute 9 for t to determine the distance fallen when t = 9 seconds. I6r 16(9) 2 1296 Substitute 16 for k. Substitute 9 for t. Simplify. Answer: 128 feet Thus, the ball falls 1,296 feet during the first 9 seconds. □ You Try It! If a 0.75 pound weight stretches a spring 5 inches, how far will a 1.2 pound weight stretch the spring? EXAMPLE 3. Tony and Paul are hanging weights on a spring in the physics lab. Each time a weight is hung, they measure the distance the spring stretches. They discover that the distance y that the spring stretches is proportional to the weight hung on the spring (Hooke's Law). If a 0.5 pound weight stretches the spring 3 inches, how far will a 0.75 pound weight stretch the spring? Solution: Let W represent the weight hung on the spring. Let y represent the distance the spring stretches. We're told that the distance y the spring stretches is proportional to the amount of weight W hung on the spring. Hence, we can write: y = kW y is proportional to W . 7.5. DIRECT AND INVERSE VARIATION 527 Substitute 3 for y, 0.5 for W, then solve for k. 3 = fc(0.5) Substitute 3 for y, 0.5 for W. 3 = k Divide both sides by 0.5. 0.5 k = 6 Simplify. Substitute 6 for k in y = kW to produce: y = 6W Substitute 6 for k in y = kW . To determine the distance the spring will stretch when 0.75 pounds are hung on the spring, substitute 0.75 for W. y = 6(0.75) Substitute 0.75 for W. y = 4.5 Simplify. Thus, the spring will stretch 4.5 inches. Answer: 8 inches □ Inversely Proportional In Examples 1, 2, and 3, where one quantity was proportional to a second quantity, you may have noticed that when one quantity increased, the second quantity also increased. Vice-versa, when one quantity decreased, the second quantity also decreased. However, not all real-world situations follow this pattern. There are times when as one quantity increases, the related quantity decreases. For example, consider the situation where you increase the number of workers on a job and note that the time to finish the job decreases. This is an example of a quantity being inversely proportional to a second quantity. Inversely proportional. We say the y is inversely proportional to x if and only if k V = -i x where k is a constant called the constant of proportionality. The phrase "y varies inversely as x" is an equivalent way of saying "y in inversely proportional to x." Here are a few examples that translate the phrase "is inversely proportional to." • Given that d is inversely proportional to t, we write d = k/t, where k is a constant. 528 CHAPTER 7. RATIONAL EXPRESSIONS • Given that y is inversely proportional to the cube of x, we write y = k/x 3 , where fc is a constant. • Given that s is inversely proportional to the square oft, we write s = k/t 2 , where k is a constant. You Try It! We are not restricted to always using the letter k for our constant of propor- tionality. Given that y is inversely EXAMPLE 4. Given that y is inversely proportional to x and the fact that proportional to x and that y = 4 when x = 2, determine the constant of proportionality, then determine y = 5 when x = 8, determine the value of y when x = 4. the value of y when x = 10. . Solution: Given the fact the y is inversely proportional to a;, we know imme- diately that k V = -, x where k is the proportionality constant. Because we are given that y = 4 when x = 2, we can substitute 4 for ?/ and 2 for a; to determine k. k x k y 4 2 8 = fc y is inversely proportional to x. Substitute 4 for y, 2 for x. Multiply both sides by 2. Substitute 8 for fc in y = k/x, then substitute 4 for x to determine y when x = 4. o y = — Substitue 8 for fc. y= 4 y = 2 Substitute 4 for x. Reduce. Answer: 4 Note that as x increased from 2 to 4, y decreased from 4 to 2. □ You Try It! If the light intensity 4 feet from a light source is 2 foot-candles, what is the intensity of the light 8 feet from the light source? EXAMPLE 5. The intensity I of light is inversely proportional to the square of the distance d from the light source. If the light intensity 5 feet from the light source is 3 foot-candles, what is the intensity of the light 15 feet from the light source? 7.5. DIRECT AND INVERSE VARIATION 529 Solution: Given the fact that the intensity / of the light is inversely propor- tional to the square of the distance d from the light source, we know immedi- ately that 7=1 d 2 ' where k is the proportionality constant. Because we are given that the intensity is J = 3 foot-candles at d = 5 feet from the light source, we can substitute 3 for / and 5 for d to determine k. I is inversely proportional to d Substitute 3 for /, 5 for d. Simplify. Multiply both sides by 25. k / - d 2 k 3 = b 2 k 3 = 2b 75 = k Substitute 75 for k in I = k/d 2 , then substitute 15 for d to determine I when d= 15. 75 d 2 75 15 2 " 75 225 1 3 Substitute 75 for k. Substitute 15 for d. Simplify. Reduce. Thus, the intensity of the light 15 feet from the light source is 1/3 foot-candle. Answer: 1/2 foot-candle □ You Try It! EXAMPLE 6. Suppose that the price per person for a camping experience is inversely proportional to the number of people who sign up for the experience. If 10 people sign up, the price per person is $350. What will be the price per person if 50 people sign up? Solution: Let p represent the price per person and let N be the number of people who sign up for the camping experience. Because we are told that the price per person is inversely proportional to the number of people who sign up for the camping experience, we can write: Suppose that the price per person for a tour is inversely proportional to the number of people who sign up for the tour. If 8 people sign up, the price per person is $70. What will be the price per person if 20 people sign up? P= N> Answer: $28 530 CHAPTER 7. RATIONAL EXPRESSIONS where k is the proportionality constant. Because we are given that the price per person is $350 when 10 people sign up, we can substitute 350 for p and 10 for N to determine k. k . . p is inversely proportional to N. p = N 350 = k 10 3500 = k Substitute 350 for p, 10 for JV. Multiply both sides by 10. Substitute 3500 for k in p = k/N, then substitute 50 for N to determine p when N = 50. p = Substitute 3500 for k. N p = Substitute 50 for TV. 1 50 p = 70 Simplify. Thus, the price per person is $70 if 50 people sign up for the camping experi- ence. □ 7.5. DIRECT AND INVERSE VARIATION 531 ti. ;». ;». Exercises ■*j ■*'. •*; 1. Given that s is proportional to £ and the fact that s = 632 when £ = 79, determine the value of s when £ = 50. 2. Given that s is proportional to £ and the fact that s = 264 when £ = 66, determine the value of s when £ = 60. 3. Given that s is proportional to the cube of £ and the fact that s = 1588867 when £ = 61, determine the value of s when £ = 63. 4. Given that d is proportional to the cube of £ and the fact that d = 318028 when £ = 43, determine the value of d when £ = 76. 5. Given that q is proportional to the square of c and the fact that q = 13448 when c = 82, determine the value of q when c = 29. 6. Given that q is proportional to the square of c and the fact that q = 3125 when c = 25, determine the value of q when c = 87. 7. Given that y is proportional to the square of x and the fact that y = 14700 when x = 70, determine the value of y when x = 45. 8. Given that y is proportional to the square of x and the fact that y = 2028 when x = 26, determine the value of y when x = 79. 9. Given that F is proportional to the cube of x and the fact that F = 214375 when x = 35, determine the value of F when x = 36. 10. Given that d is proportional to the cube of t and the fact that d = 2465195 when t = 79, determine the value of d when £ = 45. 11. Given that d is proportional to t and the fact that d = 496 when t = 62, determine the value of d when t = 60. 12. Given that d is proportional to t and the fact that d = 405 when t = 45, determine the value of d when t = 65. 13. Given that h is inversely proportional to x and the fact that h = 16 when x = 29, determine the value of h when x = 20. 14. Given that y is inversely proportional to x and the fact that y = 23 when x = 15, determine the value of y when j; = 10. 15. Given that q is inversely proportional to the square of c and the fact that q = 11 when c = 9, determine the value of q when c = 3. 16. Given that s is inversely proportional to the square of t and the fact that s = 11 when t = 8, determine the value of s when t= 10. 17. Given that F is inversely proportional to x and the fact that F = 19 when x = 22, determine the value of F when x = 16. 18. Given that d is inversely proportional to t and the fact that d = 21 when £ = 16, determine the value of d when t = 24. 532 CHAPTER 7. RATIONAL EXPRESSIONS 19. Given that y is inversely proportional to the square of x and the fact that y = 14 when x = 4, determine the value of y when x = 10. 20. Given that d is inversely proportional to the square of t and the fact that d = 21 when t = 8, determine the value of ci when t = 12. 21. Given that d is inversely proportional to the cube of t and the fact that d = 18 when £ = 2, determine the value of ci when £ = 3. 22. Given that g is inversely proportional to the cube of c and the fact that q = 10 when c = 5, determine the value of q when c = 6. 23. Given that q is inversely proportional to the cube of c and the fact that q = 16 when c = 5, determine the value of <? when c = 6. 24. Given that q is inversely proportional to the cube of c and the fact that q = 15 when c = 6, determine the value of <? when c = 2. 25. Joe and Mary are hanging weights on a spring in the physics lab. Each time a weight is hung, they measure the distance the spring stretches. They discover that the distance that the spring stretches is proportional to the weight hung on the spring. If a 2 pound weight stretches the spring 16 inches, how far will a 5 pound weight stretch the spring? 26. Liz and Denzel are hanging weights on a spring in the physics lab. Each time a weight is hung, they measure the distance the spring stretches. They discover that the distance that the spring stretches is proportional to the weight hung on the spring. If a 5 pound weight stretches the spring 12.5 inches, how far will a 12 pound weight stretch the spring? 27. The intensity / of light is inversely pro- portional to the square of the distance d from the light source. If the light inten- sity 4 feet from the light source is 20 foot- candles, what is the intensity of the light 18 feet from the light source? 28. The intensity / of light is inversely pro- portional to the square of the distance d from the light source. If the light inten- sity 5 feet from the light source is 10 foot- candles, what is the intensity of the light 10 feet from the light source? 29. Suppose that the price per person for a camping experience is inversely propor- tional to the number of people who sign up for the experience. If 18 people sign up, the price per person is $204. What will be the price per person if 35 people sign up? Round your answer to the near- est dollar. 30. Suppose that the price per person for a camping experience is inversely propor- tional to the number of people who sign up for the experience. If 17 people sign up, the price per person is $213. What will be the price per person if 27 people sign up? Round your answer to the near- est dollar. 7.5. DIRECT AND INVERSE VARIATION 533 m- j*- j*- Answers •*$ •** •** 1. 400 3. 1750329 5. 1682 7. 6075 9. 233280 11. 480 13. 116/5 15. 99 17. 209/8 19. 56/25 21. 16/3 23. 250/27 25. 40 inches 27. 1.0 foot-candles 29. $105 Chapter 8 Quadratic Functions In this chapter, we will be solving equations using the inverse operations of squaring and square roots (radicals). We will be using the familiar Pythagorean theorem to solve problems with right triangles. In addition, we will learn how use the technique of "completing the square" to develop the quadratic formula to help solve quadratic equations that are not factorable by methods of previous chapters. Four thousand years ago, the Babylonians were solving selected quadratic equations. The Greek mathematician Pythagoras did geometrical studies of which the Pythagorean theorem for right triangles is well known even in modern times. In circa 250 BC Diophantes of Alexandria and in 750 AD al Khwarizmis of Arabia also wrote quadratic equations with solutions. These earlier math- ematicians limited their work to positive rational numbers, but in medieval times the Hindu Brahmagupta included solutions using positive and negative numbers and also included irrational numbers as well. In the 1600's, Frenchman Rene Descartes wrote a geometric proof of the quadratic formula which simplified the previous information. His generation of mathematicians formulated the symbols which make it easier for us to solve problems with radicals and higher powers. 535 536 CHAPTER 8. QUADRATIC FUNCTIONS 8.1 Introduction to Radical Notation We know how to square a number. For example: • 5 2 = 25 • (-5) 2 = 25 Taking the square root of a number is the opposite of squaring. • The nonnegative square root of 25 is 5. • The negative square root of 25 is —5. Thus, when searching for a square root of a number, we are searching for number whose square is equal to our number. EXAMPLE 1. Find the square roots of 81. Solution: We are looking for a number whose square is 81. • Because 9 2 = 81, the nonnegative square root of 81 is 9. • Because (— 9) 2 = 81, the negative square root of 81 is —9. Hence, 81 has two square roots, —9 and 9. □ You Try It! Find the square roots of 64. Answer: 8 and You Try It! Find the square roots of 100. EXAMPLE 2. Find the square roots of 0. Solution: We are looking for a number whose square is 0. • Because 2 = 0, the nonnegative square root of is 0. No other number squared will equal zero. Hence, zero has exactly one square Answer: 10 and —10 root, namely zero. □ You Try It! Find the square roots of -25. EXAMPLE 3. Find the square roots of -36. .1. INTRODUCTION TO RADICAL NOTATION 537 Solution: We are looking for a number whose square is —36. However, every time you square a real number, the result is never negative. Hence, —36 has no real square roots. 1 Answer: no real square roots The introductions in Examples 1, 2, and 3 lead to the following definition. □ Defining the square roots of a number. The solutions of x 2 = a are called square roots of a. Case: a > 0. The equation x 2 = a has two real solutions, namely x = ±-y/a. • The notation y/a calls for the nonegative square root. • The notation —y/a calls for the negative square root. Case: a = 0. The equation x 2 = has exactly one solution, namely x = 0. Case: a < 0. The equation x 2 = a has no real solutions. EXAMPLE 4. Solve x 2 = 9 for x, then simplify your answers. Solution: Because the right-hand side of x 2 = 9 is positive, the equation has two solutions. You Try It! Solve x 2 = 16 for x, then simplify your answers. 9 ±V9 Original equation. Two answers: — v9 and v9 To simplify these answers, we need to understand the following facts: • V9 calls for the nonnegative square root of 9. Because (3) 2 = 9, the nonnegative square root of 9 is 3. Hence, V9 = 3. • — V9 calls for the negative square root of 9. Because (— 3) 2 = 9, the negative square root of 9 is —3. Hence, — V9 = —3. Thus, the solutions of x 2 or x = 3." 9 are x = ±3, which is equivalent to saying "x Answer: 4, □ 1 When we say that —36 has no real square roots, we mean there are no real numbers that are square roots of —36. The reason we emphasize the word real in this situation is the fact that —36 does have two square roots that are elements of the complex numbers, a set of numbers that are usually introduced in advanced courses such as college algebra or trigonometry. 538 CHAPTER 8. QUADRATIC FUNCTIONS You Try It! Solve x 2 = 49 for x, then simplify the answers. Answer: 7, EXAMPLE 5. Solve x 2 = for x, then simplify the answer. Solution: There is only one number whose square equals 0, namely 0. x 2 = x = Original equation. One answer: (0) = 0. Thus, the only solution of x 2 = is x = 0. Consequently, the nonnegative square root of zero is zero. Hence, V0 = 0. □ You Try It! Solve -9 for x, then simplify the answers. Answer: no real solutions EXAMPLE 6. Solve -4 for x, then simplify the answer. Solution: You cannot square a real number and get a negative result. Hence, -4 has no real solutions. Therefore, -4 is not a real number. □ You Try It! Simplify: -\/l44 EXAMPLE 7. Simplify each of the following: a) y/Ul b) -v/225 c) V^wo d) /324 Solution: Remember, the notation y/a calls for the nonnegative square root of a, while the notation —y/a calls for the negative square root of a. a) Because ll 2 = 121, the nonnegative square root of 121 is 11. Thus: '121 = 11 b) Because (— 15) 2 = 225, the negative square root of 225 is —15. Thus: 7 225 -15 Answer: —12 c) You cannot square a real number and get —100. Therefore, V - 100 is not a real number. d) Because (— 18) 2 = 324, the negative square root of 324 is —18. Thus: 7 324 -18 □ .1. INTRODUCTION TO RADICAL NOTATION 539 Squaring "undoes" taking the square root. Squaring square roots. If a > 0, then both —yfa and ^fa are solutions of a. Consequently, if we substitute each of them into the equation x we get: (-V5) s and W' EXAMPLE 8. Simplify each of the following expressions: a) (VE) 2 b) (-V7) 2 Solution: We'll handle each case carefully. c) (V^TT) 2 a) Because V5 is a solution of x 2 = 5, if we square V5, we should get 5. (A) 2 = 5 b) Because — \fl~ is a solution of x 2 = 7, if we square —y/7, we should get 7. (-V7) 2 7 You Try It! Simplify: (-v 7 ^) 5 c) Because x 2 = — 11 has no real answers, \J — 11 is not a real number. Ad- vanced courses such as college algebra or trigonometry will introduce the complex number system and show how to handle this expression. Answer: 21 □ Using the Graphing Calculator Up to this point, the equation x 2 = a has involved perfect squares. For exam- ple, if we start with x 2 = 25, then the solutions are x = ±v25. Because 25 is a perfect square, we can simplify further, arriving at x = ±5. However, the right-hand side of x 2 = a does not have to be a perfect square. For example, the equation x 2 = 7 has two real solutions, x = ±V7. Because 7 is not a perfect square, we cannot simplify further. In the next example, we'll use the graphing calculator to compare this algebraic solution with a graphical solution and hopefully provide some assurance that — v7 and v7 are perfectly valid solutions of ; 7. 540 CHAPTER 8. QUADRATIC FUNCTIONS You Try It! Solve the equation x 2 = 5 both algebraically and graphically, then compare your answers. EXAMPLE 9. Use the graphing calculator to solve x 2 = 7. Then solve the equation algebraically and compare answers. Solution: Enter each side of the equation x 2 = 7 in the Y= menu (see Figure 8.1), then select 6:ZStandard to produce the image in Figure 8.1. Plotl PlotE Plots \ E / WiBX A 2 \ : / -■■ViB? \ : / Wj=I \ J. We = ^Vs= \V? = Figure 8.1: Sketch each side of x 2 = 7 . Use the 5: intersect utility on the CALC menu to find the points of intersec- tion. Press ENTER in response to "First curve," press ENTER in response to "Second curve," then use the arrow keys to move the cursor closer to the point of intersection on the left than the one on the right. Press ENTER in response to "Guess." This will produce the point of intersection shown in image on the left in Figure 8.2. Repeat the procedure to find the point of intersection in the image on the right in Figure 8.2. V [ / \ \l Intersection K=-£.fiH£7£i •{=? \ [ J. \ \l Intersection K=2.fiH£7£13 •{=? Figure 8.2: Finding points of intersection. The approximate solutions of x 2 = 7 are x fs —2.645751 and x « 2.6457513. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with x and y, respectively (see Figure 8.3). • Place your WINDOW parameters at the end of each axis (see Figure 8.3). • Label each graph with its equation (see Figure 8.3). • Drop dashed vertical lines through each point of intersection. Shade and label the £- values of the points where the dashed vertical line crosses the x-axis. These are the solutions of the equation x 2 = 7 (see Figure 8.3). .1. INTRODUCTION TO RADICAL NOTATION 541 Figure 8.3: Reporting your graphical solution on your homework. Now we solve the equation algebraically. 2 n x = 7 x = ±V7 At this point, the question is: "Do these algebraic solutions match the graphical solutions in Figure 8.3?" Let's use our calculator to compare results. Locate the square root symbol \J on the calculator case above the x 2 key in the leftmost column on the calculator keyboard. Note that we will have to use the 2nd key to access this operator. Enter — <J~ (7) and press ENTER. Then enter y/~ (7) and press ENTER. The results are shown in Figure 8.4. Answer: — v5, Vo -JX7S -2.645751311 4"C7) 2.645751311 Figure 8.4: Computing — \fl and v7. Thus, -y/7 ~ -2.645751311 and V7 w 2.645751311. Note how these closely match the graphical approximations in Figure 8.3. -■J"C5> -2.236S&7977 4"<5) 2.236067977 V 1/ \ :/ Intersection K=-2.23fi0fiH V=E \ !/ \ :/ InttFS«ction X=2.23G0fiB V=E D 542 CHAPTER 8. QUADRATIC FUNCTIONS You Try It! List of Squares n n 2 1 1 2 4 3 9 4 16 5 25 6 36 7 49 8 64 9 81 10 100 11 121 12 144 13 169 14 196 15 225 16 256 17 289 18 324 19 361 20 400 21 441 22 484 23 529 24 576 25 625 EXAMPLE 10. Use the graphing calculator to solve x 2 the equation algebraically and compare answers. -5. Then solve Solution: Enter each side of the equation x 2 = —5 in the Y= menu (see Figure 8.5), then select 6:ZStandard to produce the image in Figure 8.5. Plotl PlOtE plots WiBX A 2 -■■ViB- ■5 Wj = ^Vh= We = vVs = \V? = Figure 8.5: Sketch each side of x 2 = —5. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with x and y, respectively (see Figure 8.6). • Place your WINDOW parameters at the end of each axis (see Figure 8.6). • Label each graph with its equation (see Figure 8.6). Because there are no points of intersection, the graph in Figure 8.6 informs us that the equation x 2 = — 5 has no real solutions. Now we solve the equation algebraically. Table 8.1: However, you cannot square a real number and get a negative answer. Hence, the equation x 2 = — 5 has no real solutions. This agrees completely with the graph in Figure 8.6. □ Approximating Square Roots The squares in the "List of Squares" shown in Table 8.1 are called perfect squares. Each is the square of a whole number. But not all numbers are perfect squares. For example, in the case of v24, there is no whole number whose square is equal to 24. However, this does not prevent V24 from being a perfectly good number. .1. INTRODUCTION TO RADICAL NOTATION 543 v 1 y = x \ 10 , 10 10 y = - -10 ■ Figure 8.6: Reporting your graphical solution on your homework. We can use the "List of Squares" to find decimal approximations when the radicand is not a perfect square. You Try It! EXAMPLE 11. Estimate V24 by guessing. Use a calculator to find a more Estimate: V83 accurate result and compare this result with your guess. Solution. From the "List of Squares," note that 24 lies betwen 16 and 25, so V24 will lie between 4 and 5, with V24 much closer to 5 than it is to 4. 16 + 24 4- 25 I > 4.8 Let's guess As a check, let's square 4.; 24 (4.8) 2 = (4.8)(4.8) = 23.04 Not quite 24! Clearly, V24 must be a little bit bigger than 4.8. Let's use a scientific calculator to get a better approximation. From our calculator, using the square root button, we find 24 « 4.89897948557. 544 CHAPTER 8. QUADRATIC FUNCTIONS Even though this is better than our estimate of 4.8, it is still only an ap- proximation. Our calculator was only capable of providing 11 decimal places. However, the exact decimal representation of v24 is an infinite decimal that never terminates and never establishes a pattern of repetition. Just for fun, here is a decimal approximation of V24 that is accurate to 1000 places, courtesy of http: //www. wolf ramalpha.com/. 4.8989794855663561963945681494117827839318949613133402568653851 3450192075491463005307971886620928046963718920245322837824971773 09196755146832515679024745571056578254950553531424952602105418235 40446962621357973381707264886705091208067617617878749171135693149 44872260828854054043234840367660016317961567602617940145738798726 16743161888016008874773750983290293078782900240894528962666325870 21889483627026570990088932343453262850995296636249008023132090729 18018687172335863967331332533818263813071727532210516312358732472 35822058934417670915102576710597966482011173804100128309322482347 06798820862115985796934679065105574720836593103436607820735600767 24633259464660565809954782094852720141025275395093777354012819859 11851434656929005776183028851492605205905926474151050068455119830 90852562596006129344159884850604575685241068135895720093193879959 87119508123342717309306912496416512553772738561882612744867017729 60314496926744648947590909762887695867274018394820295570465751182 126319692156620734019070649453 If you were to multiply this number by itself (square the number) , you would get a number that is extremely close to 24, but it would not be exactly 24. Answer: 9.1 There would still be a little discrepancy. □ .1. INTRODUCTION TO RADICAL NOTATION 545 ti. ;». ;». Exercises ■*j ■*-: •*; 1. List all real square roots of —400. 2. List all real square roots of 64. 3. List all real square roots of —25. 4. List all real square roots of 81. 5. List all real square roots of 49. 6. List all real square roots of —100. 7. List all real square roots of 324. 8. List all real square roots of 36. 9. List all real square roots of —225. 10. List all real square roots of 0. 11. List all real solutions of x 2 12. List all real solutions of x 2 13. List all real solutions of x 2 -225. -25. 361. 14. List all real solutions of x 2 = 256. 15. List all real solutions of x 2 = —400. 16. List all real solutions of x 2 = 0. 17. List all real solutions of x 2 = 169. 18. List all real solutions of ; -100. 19. List all real solutions of x 2 = 625. 20. List all real solutions of x 2 = 324. In Exercises 21-30, simplify each of the given expressions ^64 21. v 22. -V^529 23. - v /z 256 24. V^529 25. -v/361 26. ^^361 27. -x/T00 28. -VT96 29. \/441 30. V49 In Exercises 31-38, simplify each of the given expressions. 31. (-VTT) 2 35. (-^/29) i 32. (-\ / 3l) 2 36. (-y/M)' 33. (V59) 2 37. (779) 2 34. (V43) 2 38. (^3) 2 546 CHAPTER 8. QUADRATIC FUNCTIONS In Exercises 39-42, for each of the given equations, first use the 5:intersect utility on the CALC menu of the graphing calculator to determine the solutions. Follow the Calculator Submission Guidelines, as demonstrated in Example 9 in reporting the solution on your homework paper. Second, solve the equation algebraically, then use your calculator to find approximations of your answers and compare this second set with the first set of answers. 39. x 2 = 37 40. x 2 = 32 41. x 2 = 11 42. x 2 = 42 ;*•;*■ t*- Answers •*;•*; •*! 1. There are no real square roots. 3. There are no real square roots. 5. -7, 7 7. -18, 18 9. There are no real square roots. 11. There are no real solutions. 13. -19, 19 15. There are no real solutions. 17. -13, 13 19. -25, 25 21. 8 23. The expression is not a real number. 25. -19 27. -10 29. 21 31. 17 33. 59 35. 29 37. 79 39. x 50 > \ 1 \l i / l\ 1 \ 1 \ 1 \ 1 \ 1 \ / i / i / i J? 10 i -6.082763 i i i i i 1 -50 ^ i 10 6.082763 i i i i i ■ i 37 Answers: ±V37 w ±6.082763 41. \ i \ i \ i \ i 30 . i / i / i / i / i / l\ 1 \ l X. A / 1 10 i i i i 10 y = 11 -3.316625 3.316625 -30 Answers: ±VH « ±3.316625 ,2. SIMPLIFYING RADICAL EXPRESSIONS 547 8.2 Simplifying Radical Expressions Let's begin by comparing two mathematical expressions V^v^ =3-4 12 V9 • 16 = Vui = 12 Note that both vWHI and V9 ■ 16 equal 12. Hence, v^v 7 !^ = V9 ■ 16. Let's look at another example. VZV9 = 2-3 = 6 \AT9 = V36 = 6 Note that both v4v9 and v4 • 9 equal 6. Hence, v4V9 = v4 ■ 9. It appears that a pattern is forming, namely: \favb = vab Let's try an example on our calculator. First enter V2v3, then enter \/2 • 3 (see Figure 8.7). Note that they produce the same result. Therefore, V2v3 = y/2T3. ■J"t2>*J"C3> 2.449489743 ■J"C2*3> 2.4494S9743 Figure 8.7: Note that V2\/3 = V2 _ 3. The above discussion leads us to the following result. Multiplication property of radicals. If a > and 6 > 0, then: \fa\fb = yah You Try It! EXAMPLE 1. Simplify each of the following expressions as much as possible: Simplify: V2v8 a) ypSs/Yi b) VT2V3 c) V2VT3 Solution: In each case, use the property y/avb = \/ab. That is, multiply the two numbers under the square root sign, placing the product under a single square root. 548 CHAPTER 8. QUADRATIC FUNCTIONS Answer: 4 VsVll = V3- 11 b) VuVs = V12-3 c) \/2\/l3 = V2 ■ 13 = V33 = V36 = ^26 = 6 □ Simple Radical Form We can also use the property y/avb = \/ab in reverse to factor out a perfect square. For example: VoO = v25v2 Factor out a perfect square. = 5\/2 Simplify: %/25 = 5. The expression 5v2 is said to be in simple radical form. Like reducing a fraction to lowest terms, you should always look to factor out a perfect square when possible. Simple radical form. If possible, always factor out a perfect square. You Try It! Place v 12 in simple radical form. EXAMPLE 2. Place \/8 in simple radical form. Solution: From V8, we can factor out a perfect square, in this case V4. V 8 = v 4v 2 Factor out a perfect square. = 2\f2 Simplify: y/l = 2. Answer: 2V3 You Try It! □ Sometimes, after factoring out a perfect square, you can still factor out another perfect square. Place V200 in simple radical EXAMPLE 3. Place v 72 in simple radical form. form. . , — / — Solution: From V72, we can factor out a perfect square, in this case v 9. Factor out a perfect square. Simplify: V9 = 3. 72 = V9V8 :3V^ ,2. SIMPLIFYING RADICAL EXPRESSIONS 549 However, from v8 we can factor out another perfect square, in this case V4. = 3\/4V2 = 3-2-V2 = 6\/2 Factor out another perfect square. Simplify: y/l = 2. Multiply: 3-2 = 6. Alternate solution. We can simplify the process by noting that we can factor out V36 from V72 to start the process. V 36v 2 Factor out a perfect square. 6v2 Simplify: v 36 = 6. 72 Although the second solution is more efficient, the first solution is still math- ematically correct. The point to make here is that we must continue to factor out a perfect square whenever possible. Our answer is not in simple radical form until we can no longer factor out a perfect square. Answer: 10\/2 The Pythagorean Theorem An angle that measures 90 degrees is called a right angle. If one of the angles of a triangle is a right angle, then the triangle is called a right triangle. It is traditional to mark the right angle with a little square (see Figure 8.8). □ A b C Figure 8.8: Right triangle AABC has a right angle at vertex C . Right triangle terminology. • The longest side of the right triangle, the side directly opposite the right angle, is called the hypotenuse of the right triangle. • The remaining two sides of the right triangle are called the legs of the right triangle. 550 CHAPTER 8. QUADRATIC FUNCTIONS Proof of the Pythagorean Theorem Each side of the square in Figure 8.9 has been divided into two segments, one of length a, the other of length b. Figure 8.9: Proving the Pythagorean Theorem. We can find the total area of the square by squaring any one of the sides of the square. A = (a + b) Square a side to find area. A = a + 2ab + b Squaring a binomial pattern. Thus, the total area of the square is A = a 2 + 2ab + b 2 . A second approach to finding the area of the square is to sum the areas of the geometric parts that make up the square. We have four congruent right triangles, shaded in light red, with base a and height 6. The area of each of these triangles is found by taking one-half times the base times the height; i.e., the area of each triangles is (l/2)ab. In the interior, we have a smaller square with side c. Its area is found by squaring its side; i.e., the area of the smaller square is c 2 . The total area of the square is the sum of its parts, one smaller square and four congruent triangles. That is: 4 | -ab 2ab Adding the area of the interior square and the area of four right triangles. Simplify: 4((l/2)a6) = 2ab. The two expressions, a 2 + 2ab + b 2 and c 2 + 2ab, both represent the total area of the large square. Hence, they must be equal to one another. a + 2ab + b = c + 2ab Each side of this equation represents the area of the large square. a + b = c Subtract 2ab from both sides. ,2. SIMPLIFYING RADICAL EXPRESSIONS 551 The last equation, a 2 c 2 , is called the Pythagorean Theorem. The Pythagorean Theorem. If a and b are the legs of a right triangle and c is its hypotenuse, then: 2 i 1,2 2 a + b = c We say "The sum of the squares of the legs of a right triangle equals the square of its hypotenuse." Good hint. Note that the hypotenuse sits by itself on one side of the equation a 2 + b 2 = c 2 . The legs of the hypotenuse are on the other side. Let's put the Pythagorean Theorem to work. You Try It! EXAMPLE 4. Find the length of the missing side of the right triangle shown Find the missing side of the below. right triangle shown below. Solution: First, write out the Pythagorean Theorem, then substitute the given values in the appropriate places. a + 2 i fo\2 (4) 2 + (3) 16 + 9 25 Pythagorean Theorem. Substitute: 4 for a, 3 for b. Square: (4) 2 = 16, (3) 2 = 9. Add: 16 + 9 = 25. The equation c 2 = 25 has two real solutions, c = —5 and c = 5. However, in this situation, c represents the length of the hypotenuse and must be a positive number. Hence: c = 5 Nonnegative square root. Thus, the length of the hypotenuse is 5. Answer: 13 □ 552 CHAPTER 8. QUADRATIC FUNCTIONS You Try It! An isosceles right triangle has a hypotenuse of length 10. Find the lengths of the legs. EXAMPLE 5. An isosceles right triangle has a hypotenuse of length 8. Find the lengths of the legs. Solution: In general, an isosceles triangle is a triangle with two equal sides. In this case, an isosceles right triangle has two equal legs. We'll let x represent the length of each leg. Use the Pythagorean Theorem, substituting x for each leg and 8 for the hy- potenuse. 2 i t.2 2 a + b = c x 2 + x 2 = 8 2 2x 2 = 64 x 2 = 32 Pythagorean Theorem. Substitute: x for a, x for b, 8 for a Combine like terms: x + x = 2x. Divide both sides by 2. The equation x 2 = 32 has two real solutions, x = — v32 and x = V32- How- ever, in this situation, x represents the length of each leg and must be a positive number. Hence: x = \/32 Nonnegative square root. Remember, your final answer must be in simple radical form. We must factor out a perfect square when possible. Answer: Each leg has length 5\/2. x = V16v2 Factor out a perfect square. x = iV2 Simplify: Vl6 = 4. Thus, the length of each leg is 4V2- □ ,2. SIMPLIFYING RADICAL EXPRESSIONS 553 Applications Let's try a word problem. EXAMPLE 6. A ladder 20 feet long leans against the garage wall. If the base of the ladder is 8 feet from the garage wall, how high up the garage wall does the ladder reach? Find an exact answer, then use your calculator to round your answer to the nearest tenth of a foot. Solution: As always, we obey the Requirements for Word Problem Solutions. 1. Set up a variable dictionary. We'll create a well-marked diagram for this purpose, letting h represent the distance between the base of the garage wall and the upper tip of the ladder. You Try It! A ladder 15 feet long leans against a wall. If the base of the ladder is 6 feet from the wall, how high up the wall does the ladder reach? Use your calculator to round your answer to the nearest tenth of a foot. 2. Set up an equation. Using the Pythagorean Theorem, we can write: 64 + /i 2 3. Solve the equation. h = 20 Pythagorean Theorem. 400 Square: 8 2 = 64 and 20 2 = 400. h = 336 Subtract 64 from both sides. h = v336 h will be the nonnegative square root. h = v 16 v 21 Factor out a perfect square. h = 4V^T Simplify: VIE = 4. 4. Answer the question. The ladder reaches 4v21 feet up the wall. Using a calculator, this is about 18.3 feet, rounded to the nearest tenth of a foot. 554 CHAPTER 8. QUADRATIC FUNCTIONS 5. Look back. Understand that when we use 18.3 ft, an approximation, our solution will only check approximately. 18.3 ft Using the Pythagorean Theorem: 8 2 + 18. 3 2 = 20 2 64 + 334.89 = 400 398.89 = 400 The approximation is not perfect, but it seems close enough to accept this solution. Answer: 13.7 feet □ ,2. SIMPLIFYING RADICAL EXPRESSIONS 555 ti. ;». ;». Exercises ■*j ■*-: •*; In Exercises 1-6, simplify the given expression, writing your answer using a single square root symbol. Check the result with your graphing calculator. 1. V^Vu 2. y/2\/l 3. VT7V2 4. \/Ey/Tl 5. v^VTf 6. VT7V3 In Exercises 7-26, convert each of the given expressions to simple radical form. 7. V56 8. ^45 9. v/99 10. v 7 ^ 11. VT50 12. v^ 13. \/40 14. \/m 15. a/28 16. 7175 17. V153 18. ^/T25 19. a/50 20. a/88 21. V^8 22. vTff 23. V44 24. v^o 25. a/ToI 26. V27 In Exercises 27-34, find the length of the missing side of the right triangle. Your final answer must be in simple radical form. 27. 28. 556 CHAPTER 8. QUADRATIC FUNCTIONS 32. 33. 34. 35. In the figure below, a right triangle is in- scribed in a semicircle. What is the area of the shaded region? C 36. In the figure below, a right triangle is in- scribed in a semicircle. What is the area of the shaded region? 37. The longest leg of a right triangle is 10 feet longer than twice the length of its shorter leg. The hypotenuse is 4 feet longer than three times the length of the shorter leg. Find the lengths of all three sides of the right triangle. 38. The longest leg of a right triangle is 2 feet longer than twice the length of its shorter leg. The hypotenuse is 3 feet longer than twice the length of the shorter leg. Find the lengths of all three sides of the right triangle. 39. A ladder 19 feet long leans against the garage wall. If the base of the ladder is 5 feet from the garage wall, how high up the garage wall does the ladder reach? Use your calculator to round your answer to the nearest tenth of a foot. 40. A ladder 19 feet long leans against the garage wall. If the base of the ladder is 6 feet from the garage wall, how high up the garage wall does the ladder reach? Use your calculator to round your answer to the nearest tenth of a foot. :.2. SIMPLIFYING RADICAL EXPRESSIONS 557 m- j*- j*- Answers •*$ •** •** 1. \/65 3. ^34 5. ^85 7. 2v/l4 9. 3\/TT 11. 5\/6 13. 2i/l0 15. 2^/7 17. 3VT7 19. 5\/2 21. 3^2 23. 2\/II 25. 2v / 26 27. 2^15 29. 2^154 31. 2^37 33. 2^74 35. 25 k 7T — 8 37. 7, 24, 25 39. 18.3 feet 558 CHAPTER 8. QUADRATIC FUNCTIONS 8.3 Completing the Square In Introduction to Radical Notation on page 536, we showed how to solve equa- tions such as x 2 = 9 both algebraically and graphically. ±3 y = 9 ■10 * Note that when we take the square root of both sides of this equation, there are two answers, one negative and one positive. A perfect square is nice, but not required. Indeed, we may even have to factor out a perfect square to put our final answer in simple form. x x = ±V8 x = ±VIV2 x = ±2^2 You Try It! Readers should use their calculators to check that — 2y2 ss —2.8284 and 2y2 : 2.8284. Now, let's extend this solution technique to a broader class of equations. Solve for a;: (x + 6) 2 = 16. EXAMPLE 1. Solve for x: (x - 4) 2 9. Solution: Much like the solutions of x 2 = 9 are x = ±3, we use a similar approach on (x — 4) 2 = 9 to obtain: (a;-4) 2 = 9 x - 4 = ±3 Original equation. There are two square roots. 8.3. COMPLETING THE SQUARE 559 To complete the solution, add 4 to both sides of the equation. x = 4 ± 3 Add 3 to both sides. Note that this means that there are two answers, namely: £ = 4-3 or £ = 4 + 3 x = 1 x = 7 Check: Check each solution by substituting it into the original equation. Substitute 1 for x: Substitute 7 for x: (a;-4) 2 = 9 (x - 4) 2 = 9 (l-4) 2 = 9 (7-4) 2 = 9 (-3) 2 = 9 (3) 2 = 9 Because the last statement in each check is a true statement, both x = 1 and x = 7 are valid solutions of (x — 4) 2 = 9. Answer: —2, —10 □ In Example 1, the right-hand side of the equation (x — 4) 2 = 9 was a perfect square. However, this is not required, as the next example will show. You Try It! EXAMPLE 2. Solve for x: (x + 5) 2 = 7. Solve for x: (x - 4) : Solution: Using the same technique as in Example 1, we obtain: (x + 5) =7 Original equation. x + 5 = ±v 7 There are two square roots. To complete the solution, subtract 5 from both sides of the equation. x = — 5 ± v7 Subtract 5 from both sides. Note that this means that there are two answers, namely: x = — 5 — v7 or x = — 5 + v7 Check: Check each solution by substituting it into the original equation. Substitute —5 — V7 for x: Substitute —5 + V7 for x: ( x + 5) 2 = 7 ( x + 5) 2 = 7 ((-5 - y/7) + 5) 2 = 7 ((-5 + y/f) + 5) 2 = 7 (-V7) 2 = 7 (V7) 2 = 7 560 CHAPTER 8. QUADRATIC FUNCTIONS Answer: 4 + \/5, 4 - \/5 You Try It! Solve for x: (x + 7) 2 = 18. Because the last statement in each check is a true statement, both x = — 5— v7 and x = —5 + V7 are valid solutions of (x + 5) 2 = 7. □ Sometimes you will have to factor out a perfect square to put your answer in simple form. EXAMPLE 3. Solve for x: (x + 4) 2 = 20. Solution: Using the same technique as in Example 1, we obtain: (x + 4) = 20 Original equation. x + 4 = ±v20 There are two square roots. x + 4 = ±v 4v 5 Factor out a perfect square. x + 4 = ±2^5 Simplify: y/i=2. To complete the solution, subtract 4 from both sides of the equation. X = —4 ± 2v 5 Subtract 4 from both sides. Note that this means that there are two answers, namely: £ = -4-2^ or ^ = -4 + 2^ Check: Although it is possible to check the exact answers, let's use our cal- culator instead. First, store —4 — 2y5 in X. Next, enter the left-hand side of the equation (x + 4) 2 = 20 (see image on the left in Figure 8.10). Note that (x + 4) 2 simplifies to 20, showing that —4 — 2\/H is a solution of (x + 4) 2 = 20. In similar fashion, the solution —4 + 2v5 also checks in (x + 4) 2 = 20 (see image on the right in Figure 8.10). -4-2*J"<5)-*X -S. 472135955 (X+4> A 2 20 -4+2*J"<5)-*X .472135955 (X+4> A 2 20 Figure 8.10: Checking —4 — 2V5 and —4 + 2V5 in the equation (x + 4) z 20. Answer: -7 + 3\/2, -7-3a/2 □ 1.3. COMPLETING THE SQUARE 561 Perfect Square Trinomials Revisited Recall the squaring a binomial shortcut. Squaring a binomial. If a and b are any real numbers, then: (a±b) 2 = a 2 ± 2ab + b 2 That is, you square the first term, take the product of the first and second terms and double the result, then square the third term. Reminder examples: (x + 3) 2 =x 2 + 2(»(3) + 3 2 = x 2 + 6 s + 9 (x-8) 2 = x 2 - 2(s)(8) + 8 2 = x 2 - 16a; + 64 Because factoring is "unmultiplying," it is a simple matter to reverse the mul- tiplication process and factor these perfect square trinomials. x 2 + 6s + 9 = (s + 3) 2 16x + 64= (x Note how in each case we simply take the square root of the first and last terms. EXAMPLE 4. Factor each of the following trinomials: a) x 2 - Ylx + 36 b) x 2 + 10s + 25 c) x 2 - 34x + 289 Solution: Whenever the first and last terms of a trinomial are perfect squares, we should suspect that we have a perfect square trinomial. a) The first and third terms of x 2 — Ylx + 36 are perfect squares. Hence, we take their square roots and try: x 2 - Ylx + 36= (x-6) 2 Note that 2(cc)(6) = 12a;, which is the middle term on the left. The solution checks. b) The first and third terms of a: 2 + 10a: + 25 are perfect squares. Hence, we take their square roots and try: a; 2 + 10s + 25 = (x + 5) 2 Note that 2(s)(5) = 10s, which is the middle term on the left. The solution checks. You Try It! Factor: x 2 + 30s + 225 562 CHAPTER 8. QUADRATIC FUNCTIONS Answer: (a; + 15)^ c) The first and third terms of x 2 — 34a; + 289 are perfect squares. Hence, we take their square roots and try: x 2 - 34a; + 289 = (x - 17) 2 Note that 2(a;)(17) = 34a;, which is the middle term on the left. The solution checks. □ Completing the Square In this section we start with the binomial x 2 + bx and ask the question "What constant value should we add to x 2 + bx so that the resulting trinomial is a perfect square trinomial?" The answer lies in this procedure. Completing the square. To calculate the constant required to make x 2 + bx a perfect square trinomial: 1. Take one- half of the coefficient of x: b 2 fb\ 2. Square the result of step one: — 1 2 _b 2 4 3. Add the result of step two to x 2 + bx: 2 , b2 x + bx H 4 If you follow this process, the result will be a perfect square trinomial which will factor as follows: 2 , b 2 ( x + bx H = \ x 4 V n 2 You Try It! Given x 2 + 16a;, complete EXAMPLE 5. Given x 2 + 12x, complete the square to create a perfect the square to create a perfect square trinomial. square trinomial. o„,„ iS „„. n„ „_„ Ji , „„„,: tl ,J Solution: Compare x + 12a; with x + bx and note that 1. Take one-half of 12: 6 2. Square the result of step one: 6 2 = 36 12. 3. Add the result of step two to x 2 + Ylx: 12a; + 36 1.3. COMPLETING THE SQUARE 563 Check: Note that the first and last terms of x 2 + 12a; + 36 are perfect squares. Take the square roots of the first and last terms and factor as follows: a; 2 + 12a; + 36 = (x + 6) 2 Note that 2(a;)(6) = 12a;, so the middle term checks. Answer: x 2 + 16a; ■ 64 = (x □ You Try It! EXAMPLE 6. Given a; 2 — 3a;, complete the square to create a perfect square Given a; 2 — 5a;, complete the trinomial. Solution: Compare x 2 — 3a; with a; 2 + bx and note that 3 1. Take one-half of —3: 2. Square the result of step one: 3\ 2 _ 9 l) ~ 4 3. Add the result of step two to x 2 — 3a;: x 2 — 3x + 9 square to create a perfect square trinomial. Check: Note that the first and last terms of x 2 — 3a; + 9/4 are perfect squares. Take the square roots of the first and last terms and factor as follows: 9 x -3x + -=[x-- 4 V 2 Note that 2(x) (3/2) = 3a;, so the middle term checks. Answer: a; 2 - 5a; + 10/4 = (x - h/2f □ Solving Equations by Completing the Square Consider the following nonlinear equation. a; 2 = 2a; + 2 The standard approach is to make one side zero and factor. a; 2 - 2a; - 2 = However, one quickly realizes that there is no integer pair whose product is ac = — 2 and whose sum is b = — 2. So, what does one do in this situation? The answer is "Complete the square." 564 CHAPTER 8. QUADRATIC FUNCTIONS You Try It! Use completing the square to help solve x 2 = 3 — 6x. EXAMPLE 7. Use completing the square to help solve x 2 = 2x + 2. Solution: First, move 2x to the left-hand side of the equation, keeping the constant 2 on the right-hand side of the equation. x 2 - 2x = 2 On the left, take one-half of the coefficient of x: (l/2)(— 2) = — 1. Square the result: (— l) 2 = 1. Add this result to both sides of the equation. x 2 - 2x + 1 = 2 + 1 x 2 - 2x + 1 = 3 We can now factor the left-hand side as a perfect square trinomial. (x -l) 2 = 3 Now, as in Examples 1, 2, and 3, we can take the square root of both sides of the equation. Remember, there are two square roots. x 1 = ±^3 Finally, add 1 to both sides of the equation. x = 1±V3 Thus, the equation x 2 = 2x + 2 has two answers, x = 1 — V3 and x = 1 + v3- Check: Let's use the calculator to check the solutions. First, store 1 — v3 in X (see the image on the left in Figure 8.11). Then enter the left- and right-hand sides of the equation x 2 = 2x + 2 and compare the results (see the image on the left in Figure 8.11). In similar fashion, check the second answer 1 + V3 (see the image on the right in Figure 8.11). 1-J"<3^X .7320508076 X A 2 .5358983849 2+X+2 .5358983849 1+J"C3^X 2.732050808 7.464101615 2+X+2 7.464101615 Figure 8.11: Checking 1 — \/3 and 1 + \/3 in the equation x 2 = 2x + 2. Answer: -3 + 2^, -3-2^3 In both cases, note that the left- and right-hand sides oi x 2 = 2a;+2 produce the same result. Hence, both 1 — v3 and 1 + V3 are valid solutions of x 2 = 2x + 2. □ 3.3. COMPLETING THE SQUARE 565 EXAMPLE 8. Solve the equation x 2 — 8x — 12 = 0, both algebraically and graphically. Compare your answer from each method. Algebraic solution: First, move the constant 12 to the right-hand side of the equation. You Try It! Solve the equation x 2 + 6x + 3 = both algebraically and graphically, then compare your answers. 8a; - 12 x 2 -8x 12 Original equation. Add 12 to both sides. Take half of the coefficient of x: (l/2)(- add 16 to both sides of the equation. -4. Square: (-4) 2 = 16. Now x 2 - 8x + 16 = 12 + 16 (aj - 4) 2 = 28 x-4 = ±^28 Add 16 to both sides. Factor left-hand side. There are two square roots. Note that the answer is not in simple radical form. x - 4 = ±yfWl x - 4 = ±2^7 x = A±2Vl Factor out a perfect square. Simplify: VI = 2. Add 4 to both sides. Graphical solution: Enter the equation y = x 2 — 8x — 12 in Yl of the Y= menu (see the first image in Figure 8.12). After some experimentation, we settled on the WINDOW parameters shown in the middle image of Figure 8.12. Once you've entered these WINDOW parameters, push the GRAPH button to produce the rightmost image in Figure 8.12. Plotl PlotE Plots WiBX A 2-8*X-12 -■■Vi = Wj = \Vh= We = \V?= WINDOW Xnin= _ 5 Xnax=15 Xscl=l Ymin=-50 Vnax=5@ Vscl=16 l¥.res=W Figure 8.12: Drawing the graph of y 8x- 12. We're looking for solutions of x 2 — 8x — 12 = 0, so we need to locate where the graph of y = x 2 — 8x — 12 intercepts the x-axis. That is, we need to find the zeros of y = x 2 — 8x — 12. Select 2:zero from the CALC menu, move the cursor slightly to the left of the first ^-intercept and press ENTER in response to "Left bound." Move the cursor slightly to the right of the first x-intercept 566 CHAPTER 8. QUADRATIC FUNCTIONS and press ENTER in response to "Right bound." Leave the cursor where it sits and press ENTER in response to "Guess." The calculator responds by finding the ^-coordinate of the x-intercept, as shown in the first image in Figure 8.13. Repeat the process to find the second ^-intercept of y in the second image in Figure 8.13. 8a;— 12 shown K= -1.291503 V=0 Z*K* K=9.29i£0£fi V=0 Figure 8.13: Calculating the zeros of y 8x- 12. Reporting the solution on your homework: Duplicate the image in your calculator's viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. • Label the horizontal and vertical axes with x and y, respectively (see Figure 8.14). • Place your WINDOW parameters at the end of each axis (see Figure 8.14). • Label the graph with its equation (see Figure 8.14). • Drop dashed vertical lines through each ir-intercept. Shade and label the a;- values of the points where the dashed vertical line crosses the x-axis. These are the solutions of the equation x 2 — 8x — 12 = (see Figure 8.14). 2 -8x- 12 -1.291503 9.2915026 >• x Figure 8.14: Reporting your graphical solution on your homework. 3.3. COMPLETING THE SQUARE 567 Thus, the graphing calculator reports that the solutions of x 2 — 8a; — 12 = Answer: —3 — \f&, —3 + V& are x w -1.291503 and x « 9.2915026. Comparing exact and calculator approximations. How well do the graphing calculator solutions compare with the exact solutions, x = 4 — 2v7 and x = 4 + 2V7? After entering each in the calculator (see Figure 8.15), the comparison is excellent! -3-J"C6> -5.449489743 -3+J"C6> -.5505102572 4-2* JX7) -1.291502622 4+2* JX7) 9.291502622 K=-£.HH9H9 V=-iE"12 Figure 8.15: Approximating exact solutions x = 4 — 2v7 and a; = 4 + 2V7- 1 K= -.££05103 V=0 D 568 CHAPTER 8. QUADRATIC FUNCTIONS **. s*. j*- Exercises «•< •** * In Exercises 1-8, find all real solutions of the given equation. Place your final answers in simple radical form. 5. x 2 = -16 6. x 2 = -104 7. x 2 = 124 8. x 2 = 148 In Exercises 9-12, find all real solutions of the given equation. Place your final answers in simple radical form. 9. (x + 19) 2 = 36 11. (x + 14) 2 = 100 10. (x - 4) 2 = 400 12. (x- 15) 2 = 100 1. x 2 = 84 2. x 2 = 88 3. x 2 = 68 4. x 2 = 112 In Exercises 13-18, square each of the following binomials. 13. (x + 23) 2 16. (x-7) 2 14. (x-5) 2 17. (x-25) 2 15. (x + 11) 2 18. (x + 4) 2 In Exercises 19-24, factor each of the following trinomials. 19. x 2 + 24x + 144 22. x 2 + 8x + 16 20. x 2 - 16x + 64 23. x 2 - 20x + 100 21. x 2 -34x + 289 24. x 2 + 16x + 64 In Exercises 25-36, for each expression, complete the square to form a perfect square trinomial. Check your answer by factoring your result. Be sure to check your middle term. 25. x 2 - 20x 27. x 2 - 6x 26. x 2 - Wx 28. x 2 - 40x 8.3. COMPLETING THE SQUARE 569 29. x 2 + 20a; 33. x 2 + 15x 30. a; 2 + 26a; 34. x 2 + 25a; 31. a; 2 + 7a; 35. a; 2 - 5a; 32. a; 2 + 19a; 36. a; 2 - 3a; In Exercises 37-52, find all real solutions, if any, of the given equation. Place your final answers in simple radical form. 37. a; 2 = 18a; - 18 45. a; 2 = 16a; - 8 38. a; 2 = 12a; - 18 46. a; 2 = 10a; - 5 39. a; 2 = 16a; - 16 47. a; 2 = -18a; - 18 40. a; 2 = 12a; - 4 48. a; 2 = -lOx - 17 41. x 2 = -16a: - 4 49. a; 2 = -16x - 20 42. a; 2 = -12x- 12 50. x 2 = -16a; - 12 43. a; 2 = 18x - 9 51. x 2 = -18x-l 44. a; 2 = 16a; - 10 52. x 2 = -12a; - 8 In Exercises 53-56, solve the given equation algebraically, stating your final answers in simple radical form. Next, use the graphing calculator to solve the equation, following the technique outlined in Example 8. Use the Calculator Submission Guidelines, as demonstrated in Example 8, when reporting the solution on your homework. Compare the solutions determined by the two methods. 53. a; 2 - 2x - 17 = 55. x 2 - 6x - 3 = 54. x 2 - 4x - 14 = 56. x 2 - 4x - 16 = **■ s*< j*- Answers •*$ •** •** 1. ±2^/2T 9. -25, -13 3. ±2\/l7 11. -24, -4 5. No real solutions 13. x 2 + 46a; + 529 7. ±2^ 15. x 2 + 22x + 121 570 CHAPTER 8. QUADRATIC FUNCTIONS 17. x 2 - 50a; + 625 37.9-3^7,9 + 3^7 19. (x + 12) 2 39.8-4^3,8 + 4^/3 21. (x-17) 2 41. -8-2a/15, -8 + 2^15 23. (x-10) 2 43. 9 - 6\/2, 9 + 6\/2 25. x 2 - 20a; + 100 45.8-2714,8 + 2^/14 27. a; 2 -6a; + 9 47. -9 - 3^7, -9 + 3\/7 29. a; 2 + 20a; + 100 49. -8 - 2\/II, -8 + 2vTI 31. a; 2 + 7a; + 49/4 51. -9 - 4^5, -9 + 4^5 33. a; 2 + 15a; + 225/4 53. 1 - 3\/2, 1 + 3\/2 35. a; 2 - 5a; + 25/4 55. 3 - 2i/3, 3 + 2^3 8.4. THE QUADRATIC FORMULA 571 8.4 The Quadratic Formula We first start with the definition of a quadratic equation. Quadratic equation. A second degree polynomial equation of the form ax 2 + bx + c = 0, where a, 6, and c are any real numbers, is called a quadratic equation in x. The goal of this section is to develop a formulaic shortcut that will provide exact solutions of the quadratic equation ax 2 + bx + c = 0. We start by moving the constant term to the other side of the equation. ax + bx + c = Quadratic equation. ax + bx = —c Subtract c from both sides. In preparation for completing the square, we next divide both sides of the equation by a. 2 b x H — x = c — a a Divide both sides by a. Now we complete the square. Take one-half of the coefficient of x, then square the result. 16 6 . . / 6 \ 2 6 2 — ■ — = — when squared gives — = — r 2 a 2a H B \2a 4a? We now add 6 /(4a ) to both sides of the equation. 6 6 2 '■- a Aa 2 a 4a' 2 x 2 + - x H = 1 Add 6 2 /(4a 2 ) to both sides. On the left, we factor the perfect square trinomial. On the right, we make equivalent fractions with a common denominator. c 4a b 2 x -\ I = ■ 1 On the left, factor. On the right, 2a J a 4a 4a 2 .' . ' create equivalent tractions with a common denominator. 6\ 2 _ 4ac b 2 Ya) ~~4a~ 2 + 4a 2 x + — ) = — ^— j + -t— o Multiply numerators and denominators. 6 2 - 4ac x H = — -= — Add fractions. 2a J 4a 2 572 CHAPTER 8. QUADRATIC FUNCTIONS When we take the square root, there are two answers. b lb 2 - Aac Two square roots. When you take the square root of a fraction, you take the square root of both the numerator and denominator. b \]b 2 - Aac 2a ^4^2 b Vb 2 - Aac x-\ = ± 2a 2a Simplify: v4a 2 = 2a. b \/b 2 — Aac x = ± Subtract 6/ (2a) from both sides. 2a 2a Because both fractions have the same denominator, we can add and subtract numerators and put the answer over the common denominator. You Try It! Solve for x: x 2 - 8x + 12 = -b± Vo 2 - 4ac 2a The quadratic formula. The equation aa; 2 + bx + c equation. Its solutions are given by = is called a quadratic —b± Vo 2 - 4ac la called the quadratic formula. Whew! Fortunately, the result is a lot easier to apply than it is to develop! Let's try some examples. EXAMPLE 1. Solve for x: x 2 - 4x - 5 = 0. Solution: The integer pair 1,-5 has product ac Hence, this trinomial factors. -5 and sum b x — Ax- 5 (s + l)(a;-5) Now we can use the zero product property to write: x + 1 x or 1 £-5 = x = 5 8.4. THE QUADRATIC FORMULA 573 Thus, the solutions are x = — 1 and x = 5. Now, let's give the quadratic formula a try. First, we must compare our equation with the quadratic equation, then determine the values of a, o, and c. ax + bx + c = x 2 - 4x - 5 = Comparing equations, we see that a = 1, b = —4, and c = —5. We will now plug these numbers into the quadratic formula. First, replace each occurrence of a, 6, and c in the quadratic formula with open parentheses. x = Z^gZJ^ The quadratic formula . 2a -( )±y/( )2-4( )( ) _ , , , ... x = ; — ; Replace a, o, and c with 2 ' open parentheses. Now we can substitute: 1 for a, —4 for 6, and —5 for c. * = -M) ±V(-4) a -4(1)^5) Substitute: x for flj _ 4 for ft| and —5 for c. a; = Simplify. Exponent first, then multiplication. 4± v/36 x = — Add: 16 + 20 = 36. x = — - — Simplify: V36 = 6. Note that because of the "plus or minus" symbol, we have two answers. 4-6 4+6 x = or x = 2 2 -2 10 x ~ ~2~ x ~ T x = — 1 x = 5 Note that these answers match the answers found using the ac-test to factor the trinomial. Answer: 2, 6 □ 574 CHAPTER 8. QUADRATIC FUNCTIONS You Try It! Solve for x: x 2 + 7x = 10 EXAMPLE 2. Solve for x: x 2 = 5x + 7. Solution: The equation is nonlinear, make one side zero. x* = 5x + 7 5a; - 7 = Original equation. Nonlinear. Make one side zero. Compare x 2 — 5x — 7 = with ax 2 + bx + c = and note that a = 1, b = —5, and c = —7. Replace each occurrence of a, b, and c with open parentheses to prepare the quadratic formula for substitution. -b± \/b 2 - Aac 2o -( )±V( ) 2 ~4( )( ) 2( ) The quadratic formula. Replace a, b, and c with open parentheses. -(-5)± V / (-5) 2 -4(l)(-7) Substitute 1 for a, —5 for b, and —7 for c. Substitute: a = 1, b = — 5, c = — 7. Exponents and multiplication first. Simplify. 2(1) 5 ± V25 + 28 5± \/53 Answer: (-7 + V89)/2, (-7- 89)/2 Check: Use the calculator to check each solution (see Figure 8.16). Note that in storing (5 — v53)/2 in X, we must surround the numerator in parentheses. <5-J"<53))/2+X -1.140054945 X A 2 1.299725277 5+X+7 1.299725277 <5+JX53))/2+X 6. 140054945 X*2 37 .70027472 5+X+7 37 .70027472 I Figure 8.16: Check (5 - V53)/2 and (5 + V53)/2. In each image in Figure 8.16, after storing the solution in X, note that the left- and right-hand sides of the original equation x 2 = 5x + 7 produce the same number, verifying that our solutions are correct. □ 8.4. THE QUADRATIC FORMULA 575 In addition to placing all square roots into simple radical form, sometimes you need to reduce your answer to lowest terms. You Try It! EXAMPLE 3. Solve for x: 7.x 2 - lOx + 1 = 0. Solve for x: 3x 2 + 8x + 2 = Solution: Compare 7a; 2 — lOx + 1 = with ax 2 + bx + c = and note that a = 7, b = —10, and c = 1. Replace each occurrence of a, b, and c with open parentheses to prepare the quadratic formula for substitution. — The quadratic formula. 2a -( )±v/( ) 2 "4( )( ) Replace a, 6, and c with 2( ) open parentheses. Substitute 7 for a, —10 for 6, and 1 for c. -(-io)±v/(-io) 2 - 2(7) -4(7)(1) Substitute: 7 for a, — 10 for 6, and 1 for c 10 ± V100-28 14 10±\/72 x = Exponent, then multiplication Simplify. 14 In this case, note that we can factor out a perfect square, namely v36. 10 ± V36V2 Vl2 = \/36V2. 14 10 ±6^ Simplify: v 36 = 6 Finally, notice that both numerator and denominator are divisible by 2. 10±6v/2 2 x = Divide numerator and 14 ~n denominator by 2. 10 6A/2 2 2 x = Distribute the 2. 14 2 5±3a/2 Simplify. 576 CHAPTER 8. QUADRATIC FUNCTIONS Answer: (-4 + VlO)/3, (-4- 10)/3 Alternate simplification. Rather than dividing numerator and denominator by 2, some prefer to factor and cancel, as follows. 10 ±6^ 14 2(5 ±3^2) 2(7) 2(5 ± 3^2) W) 5±3y / 2 7 Original answer. Factor out a 2. Cancel. Simplify. Note that we get the same answer using this technique. □ You Try It! An object is launched vertically and its height y (in feet) above ground level is given by the equation y = i60 + 96i- 16i 2 , where t is the time (in seconds) that has passed since its launch. How much time must pass after the launch before the object returns to ground level? EXAMPLE 4. An object is launched vertically and its height y (in feet) above ground level is given by the equation y = 320 + 192t — 16f 2 , where t is the time (in seconds) that has passed since its launch. How much time must pass after the launch before the object returns to ground level? After placing the answer in simple form and reducing, use your calculator to round the answer to the nearest tenth of a second. Solution: When the object returns to ground level, its height y above ground level is y = feet. To find the time when this occurs, substitute y = in the formula y = 320 + 192i - 16£ 2 and solve for t. y = 320 + 192* - 16r = 320+192t- 16£ 2 Each of the coefficients is divisible by —16. = t-12t- 20 Original equation. Set y = 0. Divide both sides by —16. Compare t 2 - lit - 20 = with at 2 + bt + c = and note that a = 1, b = -12, and c = —20. Replace each occurrence of a, &, and c with open parentheses to prepare the quadratic formula for substitution. Note that we are solving for t this time, not x. -b ± Vb 2 - 4ac 2a -( )±V( ) 2 ~4( )( ) 2( ) The quadratic formula. Replace a, b, and c with open parentheses. •A. THE QUADRATIC FORMULA 577 Substitute 1 for -12 for b, and -20 for t = rs Substitute: 1 for a, I, 2(1) 12 ± V144 + 80 -12 for b, and -20 for c Exponent, then multiplication. Simplify. The answer is not in simple form, as we can factor out vl6 12± v / 16\ / 14 I t 12±4v / 14 V224 = V16V14. Simplify: VT6 = 4 Use the distributive property to divide both terms in the numerator by 2. 12 4\/li t= —± — — 2 2 t = 6±2^ Divide both terms by 2. Simplify. Thus, we have two solutions, t = 6 — 2V14 and t = 6 + 2V14. Use your calculator to find decimal approximations, then round to the nearest tenth. i« -1.5, 13.5 The negative time is irrelevant, so to the nearest tenth of a second, it takes the object approximately 13.5 seconds to return to ground level. 6-2*J"04) -1.4S33 14774 6+2*J"04) 13.48331477 Answer: 3 + v!9 w 7.4 seconds □ You Try It! EXAMPLE 5. Arnie gets on his bike at noon and begins to ride due north at a constant rate of 12 miles per hour. At 1:00 pm, Barbara gets on her bike at the same starting point and begins to ride due east at a constant rate of 8 miles per hour. At what time of the day will they be 50 miles apart (as the crow flies)? Don't worry about simple form, just report the time of day, correct to the nearest minute. Solution: At the moment they are 50 miles apart, let t represent the time that Arnie has been riding since noon. Because Barbara started at 1 pm, she has been riding for one hour less than Arnie. So, let t — 1 represent the numbers of hours that Barbara has been riding at the moment they are 50 miles apart. At 6:00 am, a freight train passes through Sagebrush Junction heading west at 40 miles per hour. At 8:00 am, a passenger train passes through the junction heading south at 60 miles per hour. At what time of the day, correct to the nearest minute, will the two trains be 180 miles apart? 578 CHAPTER 8. QUADRATIC FUNCTIONS 8(«-l) Figure 8.17: 50 miles apart. Now, if Arnie has been riding at a constant rate of 12 miles per hour for * hours, then he has traveled a distance of 12* miles. Because Barbara has been riding at a constant rate of 8 miles per hour for * — 1 hours, she has traveled a distance of 8(* — 1) miles. The distance and direction traveled by Arnie and Barbara are marked in Figure 8.17. Note that we have a right triangle, so the sides of the triangle must satisfy the Pythagorean Theorem. That is, (12t) 2 + [8{t - l)} 2 = Distribute the 8. (12i) 2 + (8t- 8) 2 = Square each term. Use (a — by 144r + 64i 2 208* 2 128* + 64 : 128* + 64 50^ 5(T 2500 2500 Use the Pythagorean Theorem. Distribute the 8. 2ab + b 2 to expand (8* - 8) 2 . Square each term. Simplify: 144* 2 + 64* 2 208r The resulting equation is nonlinear. Make one side equal to zero. 208r - 128* - 2436 = 52* 2 - 32* - 609 = Subtract 2500 from both sides. Divide both sides by 4. Compare 52i 2 - 32* - 609 = with at 2 + bt + c = and note that a = 52, b = —32, and c = —609. Replace each occurrence of a, b, and c with open parentheses to prepare the quadratic formula for substitution. Note that we are solving for * this time, not x. -b± yfb 2 - 4ac 2a -( )±V( ) 2 ~4( )( ) 2( ) The quadratic formula. Replace a, 6, and c with open parentheses. Substitute 52 for a, —32 for b, and —609 for c -(-32) ± y(-32) 2 -4(52)(-609) t 2(52) 32 ± V1024+ 126672 104 * 32 ± V127696 104 Substitute: 52 for a, -32 for 6, and -609 for c Exponent, then multiplication. Simplify. Now, as the request is for an approximate time, we won't bother with simple form and reduction, but proceed immediately to the calculator to approximate •A. THE QUADRATIC FORMULA 579 {32-JX1 27696) VI 04 -3. 12S32472 C32+J"a27696>Vl 04 3.743709336 Figure 8.18: Approximate time that Arnie has been riding. this last result (see Figure 8.18). Thus, Arnie has been riding for approxi- mately 3.743709336 hours. To change the fractional part 0.743709336 hours to minutes, multiply by 60 min/hr. 0.743709336 hr = 0.743709336 hr x 60min hr 44.62256016 min Rounding to the nearest minute, Arnie has been riding for approximately 3 hours and 45 minutes. Because Arnie started riding at noon, the time at which he and Barbara are 50 miles apart is approximately 3:45 pm. Answer: 9:42 am □ 580 CHAPTER 8. QUADRATIC FUNCTIONS f> t* t* Exercises ■** •** ■** In Exercises 1-8, solve the given equation by factoring the trinomial using the ac-method, then applying the zero product property. Secondly, craft a second solution using the quadratic formula. Compare your answers. 1. x 2 - 3x - 28 = 5. x 2 - 2x - 48 = 2. x 2 -Ax- 12 = 6. x 2 + 9a; + 8 = 3. a; 2 -8a; + 15 = 7. a; 2 + x - 30 = 4. a; 2 - 6a; + 8 = 8. a; 2 - 17a: + 72 = In Exercises 9-16, use the quadratic formula to solve the given equation. Your final answers must be reduced to lowest terms and all radical expressions must be in simple radical form. 9. a; 2 - 7x - 5 = 13. a; 2 - 7a; - 4 = 10. 3a; 2 - 3a; - 4 = 14. a; 2 - 5a; + 1 = 11. 2a; 2 +a;-4 = 15. 4a; 2 - x - 2 = 12. 2a; 2 + 7a;- 3 = 16. 5a; 2 + x - 2 = In Exercises 17-24, use the quadratic formula to solve the given equation. Your final answers must be reduced to lowest terms and all radical expressions must be in simple radical form. 17. a; 2 - x- 11 = 21. a; 2 - 3a; - 9 = 18. a; 2 - 11a; +19 = 22. a; 2 - 5a; - 5 = 19. a; 2 - 9a; + 9 = 23. a; 2 - 7a; - 19 = 20. a; 2 + 5a; - 5 = 24. a; 2 + 13a; + 4 = In Exercises 25-32, use the quadratic formula to solve the given equation. Your final answers must be reduced to lowest terms and all radical expressions must be in simple radical form. 25. 12a; 2 + 10a; - 1 = 29. 2a; 2 - 12a; + 3 = 26. 7a; 2 + 6a; - 3 = 30. 2a; 2 - 6a; - 13 = 27. 7a; 2 - 10a; + 1 = 31. 13a; 2 - 2a; - 2 = 28. 7a; 2 + 4a; - 1 = 32. 9a; 2 - 2a; - 3 = •A. THE QUADRATIC FORMULA 581 33. An object is launched vertically and its height y (in feet) above ground level is given by the equation y = 240 + 160£ — 16i 2 , where t is the time (in seconds) that has passed since its launch. How much time must pass after the launch before the object returns to ground level? Af- ter placing the answer in simple form and reducing, use your calculator to round the answer to the nearest tenth of a second. 34. An object is launched vertically and its height y (in feet) above ground level is given by the equation y = 192 + 288£ — IQt 2 , where t is the time (in seconds) that has passed since its launch. How much time must pass after the launch before the object returns to ground level? Af- ter placing the answer in simple form and reducing, use your calculator to round the answer to the nearest tenth of a second. 35. A manufacturer's revenue R accrued from selling x widgets is given by the equation R = 6000a; — 5a; 2 . The manufacturer's costs for building x widgets is given by the equation C = 500000 + 5.25a;. The break even point for the manufacturer is defined as the number of widgets built and sold so the the manufacturer's revenue and costs are identical. Find the number of widgets required to be built and sold so that the manufacturer "breaks even." Round your answers to the nearest widget. 36. A manufacturer's revenue R accrued from selling x widgets is given by the equation R = 4500a; — 15.25a; 2 . How many widgets must be sold so that the manufacturer's revenue is $125,000? Round your answers to the nearest widget. 37. Mike gets on his bike at noon and begins to ride due north at a constant rate of 6 miles per hour. At 2:00 pm, Todd gets on his bike at the same starting point and be- gins to ride due east at a constant rate of 8 miles per hour. At what time of the day will they be 60 miles apart (as the crow flies)? Don't worry about simple form, just report the time of day, correct to the nearest minute. 38. Mikaela gets on her bike at noon and be- gins to ride due north at a constant rate of 4 miles per hour. At 1:00 pm, Rosemarie gets on her bike at the same starting point and begins to ride due east at a constant rate of 6 miles per hour. At what time of the day will they be 20 miles apart (as the crow flies)? Don't worry about simple form, just report the time of day, correct to the nearest minute. 39. The area of a rectangular field is 76 square feet. The length of the field is 7 feet longer than its width. Find the dimensions of the field, correct to the nearest tenth of a foot. 40. The area of a rectangular field is 50 square feet. The length of the field is 8 feet longer than its width. Find the dimensions of the field, correct to the nearest tenth of a foot. 582 CHAPTER 8. QUADRATIC FUNCTIONS 41. Mean concentrations of carbon dioxide over Mauna Loa, Hawaii, are gathered by the Earth System Research Laboratory (ESRL) in conjunction with the National Oceacnic and Atmosphere Administration (NOAA). Mean annual concentrations in parts per million for the years 1962, 1982, and 2002 are shown in the following table. Year Concentration (ppm) A quadratic model is fitted to this data, yielding 1962 1982 2002 318 341 373 C = 0.01125r + 0.925£ + 318, where t is the number of years since 1962 and C is the mean annual concentration (in parts per million) of carbon dioxide over Manuna Loa. Use the model to find the year when the mean concentration of carbon dioxide was 330 parts per million. Round your answer to the nearest year. 42. The U.S. Census Bureau provides historical data on the number of Americans over the age of 85. Year Population over 85 (millions) A quadratic model is fitted to this data, yielding 1970 1990 2010 1.4 3.0 5.7 P = 0.01375i 2 + 0.0525* + 1.4, where t is the number of years since 1970 and P is number of Americans (in millions) over the age of 85. Use the model to find the year when the number of Americans over the age of 85 was 2,200,000. Round your answer to the nearest year. j*- j*- j*- Answers •*$ **s ■*$ 1.-4,7 ,l±v^$3 15. 17. 3. 3, 5 5. -6, 8 7. -6, 5 19 1±3V5 2 9 ±3^ 2 7± ^9 3±3V5 9 2 2 -1±a/33 7 ±5^5 U * —^~ 23 - —2— 13. l±M 25 . - 5± ^ 2 12 8.4. THE QUADRATIC FORMULA 583 5_±_3_y/2 35. 90 widgets, 1109 widgets 7 6±V30 37 - 7:12 pm 2 ^- 39. 5.9 by 12.9 feet 31. — 13 41. 1973 33. 11.3 seconds 27. 29. Index aomethod, 414 graphing calculator, 462 revised, 417 ^-intercept, 234 ^-intercept and zero of a function, 325 y-intercept, 201, 234 absolue value, 20 absolute value, 4, 16 adding fractions, 36 adding integers, 5 additive identity property, 2 algebraic expression, 21 algebraic expressions, 60 applications, 121 approximating square roots, 542 area rectangle, 471 area of parallelogram, 119 area of rectangle, 118 area of trapezoid, 118 area of triangle, 116 ascending powers, 309 associative property of addition, 7 associative property of multiplica- tion, 9, 60 binomial, 309 brackets, 16 calculator submission guidelines, 175, 408, 419, 432, 520, 566 canceling, 100 cancellation, 34 Cartesian Coordinate System, 152 circumference of circle, 119 clearing decimals, 99, 106 clearing fractions, 99, 101, 113, 516 clearing negative exponents, 485 coefficient, 308 combining like terms, 64 commutative property of addition, 6 commutative property of multipli- cation, 9 complementary angles, 282 completing the square, 558 procedure, 562 solving equations, 563 consecutive even integers, 124, 470 consecutive integers, 122 constant of proportionality, 525, 527 constant rate, 184 cost, 339 decimals, 48 addition, 49 division, 51 multiplication, 51 rounding, 55 subtraction, 49 degree, 308, 311 demand, 363 dependent variable, 183 585 586 INDEX descending powers, 309 difference of squares, 373, 444 distributing a negative sign, 63 distributive property, 61 distributive property for division, 511 dividing fractions, 35 divisor, 386 domain, 294 elimination method, 273 equation, 74 equivalent, 75 solution, 74 equations clearing decimals, 99, 106 clearing fractions, 99, 101, 113 different from expressions, 74 equivalent equations, 75 isolate terms containing x, 91 meaning of "solve for x" , 90, 111 operations producing equivalent equations, 77, 79, 80 simplify using order of opera- tions, 92 equations in two variables, 156 solution, 157 equations with more than one vari- able, 111 equivalent equations, 75 adding same amount to both sides, 77 dividing both sides by same amount, 80 multiplying both sides by same amount, 79 subtracting same amount to both sides, 77 equivalent inequalities, 139 evaluating algebraic expressions, 21 exponents clearing negative exponents, 485 definition, 10, 343 dividing like bases, 481 multiplying like bases, 480 negative, 478 raising a power to a power, 481 raising to —1, 478 raising to a negative integer, 482 raising to negative integer, 479 factor, 386 factoring ax 2 + bx + c, a = 1, 413 ax 2 +bx + c,a^l, 427 applications, 467 by grouping, 394 checking the solution, 457 definition, 390 difference of squares, 444 factor out GCF, 390 first rule of factor, 430 first rule of factoring, 390, 443, 446 perfect square trinomials, 442, 561 special forms, 440 strategy, 456 factoring completely, 446, 456 FOIL method, 370, 372 formulae, 111 fractions addition, 36 canceling, 100 division, 35 multiplication, 33 free fall, 526 function definition, 297 rule of three, 301 function notation, 301 functions, 294 y and f(x) are equivalent, 303 adding, 334 sketching the graph, 304 subtracting, 337 using different letters for dif- ferent functions, 303 zero, 325 geometric formula INDEX 587 sum of angle of a triangle, 126 geometric formulae, 115 area of a circle, 362 area of parallelogram, 119 area of rectangle, 118 area of trapezoid, 118 area of triangle, 116 circumference of circle, 119 perimeter of rectangle, 115 graph of an equation, 157 graphing scaling each axis, 183 graphing an equation guidelines, 160 graphing calculator, 171 ac-method, 462 adjusting the viewing window, 176 adjusting WINDOW parame- ters, 256 changing system solutions to frac- tions, 267 check solutions of rational equa- tion, 518 checking fractional solutions, 105 checking solutions, 78, 92 checking solutions of equations, 89 checking solutions of systems, 264 checking solutions to nonlinear equation, 403 checking solutions with radicals, 560, 574 drawing parabola, 314 drawing polynomial of degree four, 316 evaluating expressions, 23 evaluating functions, 302 evaluating polynomial function, 323 finding zero of a function, 326 fractions, 39 intersect utility, 253, 406, 418, 432, 449, 540 mode button, 171 negation versus subtraction, 11 plotting equation, 172 reproducing results on home- work, 174 rounding decimals, 55 scientific notation, 497 solving x 2 = a, 540 solving nonlinear equations, 405 solving rational equations, 519 solving systems, 253 STAT PLOT, 172 submissioin guidelines, 175 table feature, 160 TRACE button, 178 WINDOW menu, 176 zero utility, 326, 407, 421, 435, 565 graphing equations by hand, 152 graphing equations in two variables, 157 greater than, 136 greater than or equal to, 136 greatest common divisor, 31, 386, 387 greatest common factor, 386, 387 two monomials, 389 grouping symbols, 16, 19 nested, 20 guidelines for graphing an equation, 160 Hooke's Law, 112, 526 horizontal line, 237 hypotentuse, 549 independent variable, 183 inequalities, 135 add or subtract same amount, 140 clearing decimals, 145 clearing fractions, 144 multiply or divide by positive number, 140 multiplying or dividing by neg- ative number, 142 reversing inequality sign, 141 588 INDEX integers, 2, 3 addition, 5 multiplication, 8 subtraction, 7 intercept, 201 intercepts, 234 interest, 286 intersect utility, 406, 418, 432, 449, 540 interval notation, 138 eyes "left to right" , 139 summary table, 146 inverse of addition, 76 inverse of division, 80 inverse of multiplication, 80 inverse of subtraction, 76 irrational numbers, 135 kinetic energy, 114 laws of exponents, 343, 480, 484 dividing like bases, 346 multiplying like bases, 344 raising a power to a power, 348 raising a product to a power, 349 raising a quotient to a power, 351 least common denominator, 36, 508 procedure for finding, 510 legs, 549 less than, 136 less than or equal to, 136 light intensity, 528 light-year, 500 like terms, 64, 65 linear equations strategy for solving, 401 linear equations versus nonlinear equa- tions, 400 list of squares, 543 lowest terms, 31 mapping diagram, 296 notation, 299 mass on a spring, 526 measuring change in variable, 185 mixture, 288 monomial, 308 multiplicative identity property, 10 multiplying fractions, 33 multiplying integers, 8 multiplying radicals, 547 natural numbers, 2 negating a sum, 336 negative exponents, 478 nested grouping symbols, 20 Newton's Law of Gravitation, 114 Newton's law of gravitation, 500 nonlinear equaitions graphing calculator solution, 404 nonlinear equations, 399 graphing calculator solution, 417, 431, 447 solving by factoring, 404, 417, 431, 447 strategy for solving, 402 nonlinear equations versus linear equa- tions, 400 number line, 135 numbers integers, 3 irrational, 135 natural, 2 rational, 31, 135 real, 135 whole, 2 Ohm's Law, 113 order of operations, 16, 37, 52, 66 rules, 16, 37, 52, 66 ordered pair, 152 ordered pairs plotting, 154 ordering real numbers, 136 origin, 153 parabola, 313 opens downward, 314 opens upward, 314 vertex, 314 INDEX 589 parentheses, 16 perfect square trinomial, 442 perimeter, 283 rectangle, 471 perimeter of rectangle, 115 perpendicular, 235 perpendicular lines, 219 plotting ordered pairs, 154 point-slope form, 213, 214 applications, 222 determine equation of line through two points, 214 parallel lines, 217 perpendicular lines, 220 to standard form, 232 polynomial, 309 ascending powers, 309 binomial, 309 coefficient, 308 degree, 308, 311 degree four, 316 graph, 313 monomial, 308 quadratic, 313 reading the graph, 321 term, 308 trinomial, 309 polynomial function, 312 evaluating, 321 polynomials, 308 adding, 334 adding and subtracting, 334 applications, 321 applications of addition and sub- traction, 338 applications of multiplication, 362 descending powers, 309 distribution shortcut, 360 division of polynomial by a mono- mial, 511 monomial products, 356 multiplication, 356 multiplying a monomial and poly- nomial, 357 multiplying polynomials, 359 negation, 335 subtracting, 337 prime factorization, 387, 510 finding greatest common divi- sor, 388 prime number, 32 projectile motion, 323, 328, 468, 576 property additive identity, 2 associative, 7, 9, 60 commutative, 6, 9 distributive, 61 multiplicative identity, 10 proportional, 525 proportionality constant, 525, 527 Pythagorean Theorem, 549, 551, 577 applications, 553 proof, 550 quadrants, 153 quadratic equation, 571 quadratic formula, 571, 572 quadratic polynomial, 313 radical expressions simplifying, 547 radical notation, 536 range, 294 rate, 183 constant, 184 rational equation clearing fractions, 516 rational equations applications, 521 checking solution with graph- ing calculator, 518 solving, 516 solving with graphing calcula- tor, 519 rational expression, 505 adding, 507 dividing, 506 least common denominator, 508 multiplying, 505 subtracting, 508 rational numbers, 31, 135 590 INDEX real numbers, 135 ordering, 136 reciprocal, 35, 521 reducing fractions, 31 relation, 294 domain, 294 range, 294 Rene Descartes, 152 reproducing calculator results on home- work, 174 requirements for word problem so- lutions, 121 revenue, 339, 363 right angle, 549 right triangle hypotenuse, 549 legs, 549 rounding, 55 rule of three, 301 scaling each axis, 183 scientific notation, 492 form definition, 495 graphing calculator, 497 how to report your answer, 499 multiply decimal numbers by negative power of ten, 494 multiply decimal numbers by nonnegative power of ten, 493 negative powers of ten, 492 nonnegative powers of ten, 492 placing a number in scientific notation, 495 set-builder notation, 136 summary table, 146 simple radical form, 548 factor out perfect square, 548 simplifying radical expressions, 547 slope, 183 determining slope from graph of the line, 193 downhill, 190 drawing a line with given slope, 194 horizontal line, 191 independent of direction of sub- traction, 189 independent of selected points, 187 parallel lines, 217 perpendicular lines, 220 steepness of line, 189 summary table, 195 uphill, 189 vertical line, 191 slope as rate, 185 slope of line through two points, 188 slope-intercept form, 201, 202 applications, 206 determine equation from graph, 205 sketching line from given equa- tion, 202 to standard form, 230 solution of an equation, 74 solution of equation in two variables, 157 solution of linear system, 246 solve for x, 111 solving equations completing the square, 563 multiple steps, 87 one step, 74 solving rational equations, 516 special product FOIL method, 372 special products, 370 difference of squares, 373 FOIL method, 370 squaring a binomial, 375, 378 speed of light, 501 square root negative, 537 nonnegative, 537 square roots approximations, 542 definition, 537 list of squares, 543 multiplication property, 547 simple radical form, 548 INDEX 591 squaring a binomial, 375, 378, 440, 561 standard form, 228 definition, 230 identifying intercepts, 234 to slope-intercept form, 228 substitution method, 262, 263 subtracting integers, 7 sum of angles of a triangle, 126 systems applications, 282 checking solution with graph- ing calculator, 264 infinite number of solutions, 252 infinite solutions, 269, 277 interpretation tip, 269 no solutions, 250, 268, 276 number of solutions, 253 parallel lines, 250 same lines, 252 solution, 246 solving by elimination, 273 solving by graphing, 246 solving by substitution, 262, 263 solving using graphing calcula- tor, 253 inverse, 527 vertex of parabola, 314 vertical line, 237 whole numbers, 2 word problem investment, 127 word problems, 121 requirements, 121 wrap and unwrap, 75, 87 writing mathematics, 17, 82 zero of a function, 325 zero product property, 399 zero utility, 407, 421, 435, 565 zeros and ^-intercepts, 406 table feature of graphing calculator, 160 term, 308 tips for evaluating algebraic expres- sions, 21 tips for using function notation, 302 trinomial, 309 unit price, 363 unlike terms, 64 variable, 21 change, 184 dependent, 183 independent, 183 maintaining case, 111 variable part, 64 variation, 525 direct, 525